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ANALYTIC    GEOMETRY 


FOR 


COLLEGES,  UNIVERSITIES,  AND 
TECHNICAL   SCHOOLS. 


BY 

E.    W.    NICHOLS, 

PROFESSOR  OF   MATHEMATICS   IN   THE   VIRGINIA   MILITARY 
INSTITUTE. 


CHESTNUT  HILL,,  MAS.S. 
MATH.  DEPT. 


LEACH,    SHEWELL,    &   SANBORN, 

BOSTON.  NEW  YORK.  CHICAGO. 


Copyright,  1892, 
By  Leach,  Shewell,  &  Sanborn, 


C.  J.   PETERS  &  SON, 
Typographers  and  Electrotypers. 


Press  of  Berwick  &  Smith. 


150580 


PREFACE 


This  text-book  is  designed  for  Colleges,  Universities, 
and  Technical  Schools.  The  aim  of  the  author  has  been 
to  prepare  a  work  for  beginners,  and  at  the  same  time  to 
make  it  sufficiently  comprehensive  for  the  requirements  of 
the  usual  undergraduate  course.  For  the  methods  of  develop- 
ment of  the  various  principles  he  has  drawn  largely  upon  his 
experience  in  the  class-room.  In  the  preparation  of  the  work 
all  authors,  home  and  foreign,  whose  works  were  available, 
have  been  freely  consulted. 

In  the  first  few  chapters  elementary  examples  follow  the 
discussion  of  each  principle.  In  the  subsequent  chapters 
sets  of  examples  appear  at  intervals  throughout  each  chapter, 
and  are  so  arranged  as  to  partake  both  of  the  nature  of  a 
review  and  an  extension  of  the  preceding  principles.  At  the 
end  of  each  chapter  general  exam,ples,  involving  a  more 
extended  application  of  the  principles  deduced,  are  placed  for 
the  benefit  of  those  who  may  desire  a  higher  course  in  the 
subject. 

The  author  takes  pleasure  in  calling  attention  to  a  "Dis- 
cussion of  Surfaces,"  by  A.  L.  Nelson,  M.A.,  Professor  of 
Mathematics  in  Washington  and  Lee  University,  which 
appears  as  the  final  chapter  in  this  work. 

He  takes  pleasure  also  in  acknowledging  his  indebtedness 


IV  PREFACE. 

to  Prof.  C.  S.  Venable,  LL.D.,  University  of  Virginia,  to 
Prof.  William  Cain,  C.E.,  University  of  North  Carolina, 
and  to  Prof.  E.  S.  Crawley,  B.S.,  University  of  Pennsylvania, 
for  assistance  rendered  in  reading  and  revising  manuscript, 
and  for  valuable  suggestions  given. 

E.  W.  Nichols. 
Lexington,  Va. 

January,  1893. 


CONTENTS, 


PAET  L— PLANE  ANALYTIC  GEOMETEY. 


CHAPTEE  I. 

co-okdinates. 
Arts.  Pages 

1-3.     The  Cartesian  or  Bilinear  System.     Examples     ....  1 

4-6.     The  Polar  System.     Examples 4 

CHAPTER   11. 

LOCI. 


7.  Locus  of  an  Equation.     The  Equation  of  a  Locus 

8.  Yariables.     Constants.     Examples 

9.  Relationship  between  a  Locus  and  its  Equation  , 
10-16.  Discussion  and  Construction  of  Loci.  Examples 
17, 18.  Methods  of  Procedure.     Examples 


10 
11 
11 
23 


CHAPTEE   III. 

THE   STRAIGHT   LINE. 

19.  The  Slope  Equation.     Examples 25 

20.  The  Symmetrical  Equation.     Examples 29 

21.  The  Normal  Equation 32 

22.  Perpendicular  Distance  of   a  Point  from   a   Line.      Ex- 

amples        33 

23.  Equation  of  Line,  Axes  Oblique.     Examples 35 

24.  General  Equation,  Ax  +  By  +  C  =  0 37 

25.  Equation  of  Line  passing  through  a  Point.     Examples    .  38 

26.  Equation   of  Line  passing  through   Two   Points.      Ex- 

amples        39 

27.  Length  of  Line  joining  Two  Points.     Examples      ...  41 


vi  CONTENTS. 

Akts.  Pages 

28.  Intersection  of  Two  Lines.     Examples 42 

29.  Ax  +  B?/  +  C  +  K  (A'x  +  B'y  +  C')  =  0 43 

30.  Angle  between    Two  Lines.      Examples.      General    Ex- 

amples           .....  44 

CHAPTER   IV. 

TRANSFOEMATION    OF    CO-OEDINATES. 

31.  Objects  of.    Illustration 50 

32.  From  One  System  to  a  Parallel  System.     Examples      .     .  51 

33.  Eectangular  System  to  an  Oblique  System.     Rectangular 

System  to  Another  System  also  Rectangular.    Examples.  53 

34, 35.  Rectangular  System  to  a  Polar  System.  From  a  Polar 
System  to  a  Rectangular  System.  Examples.  General 
Examples 55 

CHAPTER  V. 

THE   CIRCLE. 

36, 37.     Generation  of  Circle.     Equation  of  Circle  ......  59 

38.  General   Equation   of    Circle.     Concentric    Circles.     Ex- 

amples        61 

39.  Polar  Equation  of  Circle 63 

40.  Supplemental  Chords 64 

41.  Tangent.     Sub-tangent 66 

42.  Normal.     Sub-normal 67 

43.  General  Equations  of  Tangent  and  Normal.     Examples    .  68 

44.  Length  of  Tangent 70 

45, 46.     Radical  Axis.     Radical  Centre.     Examples 70 

47.  Condition  that  a  Straight   Line   touch  a  Circle.      Slope 

Equation  of  Tangent 74 

48.  Chord  of  Contact 74 

49, 50.     Pole  and  Polar 75 

51.     Conjugate  Diameters.     Examples.     General  Examples     .  77 

CHAPTER    VI. 

THE   PAKABOLA. 

52,  53.  Generation  of  Parabola.  Equation  of  Parabola.  Defini- 
tions   83 

54.  Construction  of  Parabola 85 

55.  Latus-Eectum.     Examples 86 


CONTENTS.  vii 

Arts.  Pages 

56.     Polar  Equation  of  Parabola 88 

57-59.     Tangent.     Sub-tangent.     Construction  of  Tangent      .     .  89 

60, 61.     Normal.    Sub-normal 90 

62.  Tangents  at  Extremities  of  Latus-Rectum 91 

63.  x^  +  yi  —  +  a^-     Examples 92 

64.  Tangent  Line  makes  Equal  Angles  with  Focal  Lines  and 

Axis 95 

65.  Condition  that  a  Straight  Line  touch  the  Parabola.     Slope 

Equation  of  Tangent 95 

66.  Locus  of  Intersection  of  Tangent  and  Perpendicular  from 

Focus 96 

67.  Locus  of  Intersection  of  Perpendicular  Tangents    ...  97 

68.  Chord  of  Contact 97 

69.  Pole  and  Polar ; 98 

70.  Conjugate  Diameters 98 

71, 72.     Parameter  of  any  Diameter.     Equation  of  a  Diameter      .  100 

73.     Tangents  at  the   Extremities    of  a   Chord.      Examples. 

General  Examples 101 

CHAPTER  VIL 

THE    ELLIPSE. 

74,  75.     Generation  of  Ellipse.    Definition.    Equation  of  Ellipse    .  106 

76, 77.     Eccentricity.     Focal  Radii 108 

78.  Construction  of  Ellipse 109 

79.  Latus-Rectum.     Examples Ill 

80.  Polar  Equation  of  Ellipse 113 

81.  Supplemental  Chords 114 

82,83.     Tangent.     Sub-tangent 115 

84.  Tangent    and    Line    through    Point    of    Tangency   and 

Centre 118 

85.  Methods  of  constructing  Tangents 118 

86, 87.     Normal.     Sub-normal.     Examples 119 

88.  Normal  bisects  Angle  between  the  Focal  Radii    ....  122 

89.  Condition  that  a  Straight  Line  touch  the  Ellipse.     Slope 

Equation  of  Tangent 123 

90.  Locus   of    Intersection  of    Tangent  and    Perpendicular 

from  Focus 124 

91.  Locus  of  Intersection  of  Perpendicular  Tangents     .     .     .  125 

92.  Equation  of  Chord  of  Contact 125 

93.  Pole  and  Polar 126 

94.  Conjugate  Diameters 126 


vm  CONTENTS. 

Akts.  Pages 
95, 96.     Equation  of  a  Diameter.     Co-ordinates  of  Extremities  of 

Conjugate  Diameter 129 

97.  a'2  +  6'^  =  a2  +  &2 230 

98.  Parallelogram  on  a  pair  of  Conjugate  Diameters      .     ,     .  131 

99.  Kelation   between   Ordinates   of   Ellipse   and  Circles  on 

Axes 133 

100, 101.     Construction  of  Ellipse.     Area  of  Ellipse.     Examples. 

Genei'al  Examples 134 

CHAPTER   VIII. 

THE   HYPERBOLA. 

102, 103.     Generation  of  Hyperbola.     Definitions.      Equation  of 

Hyperbola 141 

104,  105.     Eccentricity.     Focal  Radii 143 

106, 107.     Construction  of  Hyperbola.     Latus-Rectum     ....  144 

108.  Relation  between  Ellipse  and  Hyperbola 146 

109.  Conjugate  Hyperbola.     Examples 146 

110.  Polar  Equation  of  Hyperbola 149 

111.  Supplemental  Chords 150 

112,113.     Tangent.     Sub-tangent 150 

114.  Tangent    and    Line    through  Point  of   Tangency  and 

Centre 151 

115.  Method  of  constructing  Tangents    ........  151 

116, 117.     Normal.     Sub-normal.     Examples 152 

118.  Tangent  bisects  Angle  between  the  Focal  Radii    .     .     .  154 

119.  Condition  that  a  Straight  Line  touch  the  Hyperbola. 

Slope  Equation  of  Tangent 155 

120.  Locus  of  intersection  of   Tangent   and   Perpendicular 

through  Focus 155 

121.  Locus  of  intersection  of  Perpendicular  Tangents      .     .  155 

122.  Chord  of  Contact 156 

123.  Pole  and  Polars 156 

124.  Conjugate  Diameters 156 

125.  Conjugate  Diameters  lie  in  the  same  Quadrant     .     .     .  157 
126, 127.     Equation   of    Conjugate  Diameter.       Co-ordinates   of 

Extremities  of  Conjugate  Diameter 157 

128.  a/2  —  6'2  =  a2  —  5'2 158 

129.  Parallelogram  on  a  pair  of  Conjugate  Diameters.     Ex- 

amples   159 

1.30.     Asymptotes 160 

131.     Asymptotes  as  Axes.      Rhombus    on    Co-ordinates  of 

Vertex 162 


CONTENTS.  ix 

Arts.                                              '  Pages 

132.  Tangent  Line,  Asymptotes  being  Axes.     The  Point 

of  Tangency 164 

133.  Intercepts  of  a  Tangent  on  the  Asymptotes      ....  165 

134.  Triangle  formed  by  a  Tangent  and  the  Asymptotes  .     .  165 

135.  Intercepts    of    a    Cliord   between    Hyperbola   and  its 

Asymptotes.     Examples.     General  Examples   .     .     .  165 

CHAPTEE   IX. 

GENERAL  EQUATION   OF  THE   SECOND   DEGREE. 

136, 187.     The  General  Equation.     Discussion 170 

138.  First  Transformation.     Signs  of  Constants      ....  171 

139.  Second  Transformation 172 

140,141.     a'=:0.     c'  =  0 178 

142.  Summary 175 

143.  lfi<4.ac ~   .     .  175 

144.  62  —  4  (jc YIQ 

145.  62  ->  4  (tc Yin 

146.  General  Summary.     Examples 178 

CHAPTER  X. 

HIGHER  PLANE   CURVES. 

147.  Definition 188 

EQUATIONS   OF  THE   THIRD   DEGREE.' 

148.  The  Semi-cubic  Parabola 188 

149.  Duplication  of  Cube  by  aid  of  Parabola 190 

150.  The  Cissoid 191 

151.  Duplication  of  Cube  by  aid  of  Cissoid 193 

152.  The  Witch 194 

EQUATIONS   OF   THE   FOURTH  DEGREE. 

153.  The  Conchoid 196 

154.  Trisection  of  an  Angle  by  aid  of  Conchoid 198 

155.  The  LimaQon 199 

156.  The  Lemniscate 201 

TRANSCENDENTAL   EQUATIONS. 

157.  The  Curve  of  Sines 208 

158.  The  Curve  of  Tangents 204 

159.  The  Cycloid 206 


X  CONTENTS. 

Arts.  Pages 

SPIRALS. 

160.  Definition 208 

161.  Tlie  Spiral  of  Arcliimedes 208 

162.  Tlie  Hyperbolic  Spiral 210 

163.  The  Parabolic  Spiral 212 

164.  The  Lituus  . 213 

165.  The  Logarithmic  Spiral.     Examples .  214 


PART   II.  — SOLID   ANALYTIC   GEOMETEY. 
CHAPTER  I. 

CO-ORDINATES. 

166.  The  Tri-planar  System.     Examples 217 

167.  Projections 219 

168, 169.     Length    of    Line    joining    Two   Points.     Directional 

Angles.     Examples 220 

170.  The  Polar  System 222 

171.  Relation    between    Systems.     Transformation  of    Co- 

ordinates.    Examples 223 

CHAPTER   II. 

THE   PLANE. 

172.  Equation  of  Plane 226 

173.  Normal  Equation  of  Plane 227 

174.  Symmetrical  Equation  of  Plane 229 

175.  General  Equation  of  Plane •  229 

176.  Traces.     Intercepts 230 

177.  Perpendicular  from  Point  on  Plane 231 

178.  Plane  through  three  Points 232 

179.  Any  Equation   between   three  Variables.     Discussion. 

Examples 283 

CHAPTER   in. 

THE   STRAIGHT    LINE. 

180.  Equations  of  a  Straight  Line 236 

181.  Symmetrical  Equations  of  a  Straight  Line 237 


CONTENTS.  xi 

Arts.  Pages 

182.  To   find  where  a   given  Line  pierces  the  Co-ordinate 

Planes " 2.38 

183.  Line  through  One  Point 239 

184.  Line  through  Two  Points.     Examples 239 

185.  Intersecting  Lines 242 

186, 187.     Angle  between  Two  Lines 243 

188.     Angle  between  Line  and  Plane 246 

189, 190.     Transformation  of  Co-oi'dinates 247 

191-193.     The  Cone  and  its  Sections 250 

194, 195.    Definitions.     Equation  of  a  Conic.     Examples    .     .     .  254 

CHAPTER  IV. 

DISCUSSION   OF   SURFACES   OF   THE    SECOND   OEDER. 

General  Equation  of  the  Second  Degree  involving  three 

Variables.    Transformations  and  Discussion       .     .     .  259 

The  Ellipsoid  and  varieties 262 

The  Hyperboloid  of  One  Sheet  and  varieties    ....  265 

The  Hyperboloid  of  Two  Sheets  and  varieties      .     .     .  267 

The  Paraboloid  and  varieties 269 

Surfaces  of  Revolution.     Examples 273 


PLANE  ANALYTIC   GEOMETRY. 


PART    I. 


CHAPTER   I. 

CO-ORDINATES.  — THE   CARTESIAN   OR  BILINEAR 
SYSTEM. 

1.  The  relative  positions  of  objects  are  determined  by 
referring  them  to  some  other  objects  whose  positions  are 
assumed  as  known.  Thus  we  speak  of  Boston  as  situated 
in  latitude  —  °  north,  and  longitude  —  °  west.  Here  tlie  ob- 
jects to  which  Boston  is  referred  are  the  equator  and  the 
meridian  passing  through  Greenwich.  Or,  we  speak  of  Bos- 
ton as  being  so  many  miles  north-east  of  New  York.  Here  the 
objects  of  reference  are  the  m^eridian  of  longitude  through 
New  York  and  Hew  York  itself.  In  the  first  case  it  will  be 
observed,  Boston  is  referred  to  two  lines  which  intersect  each 
other  at  right  angles,  and  the  position  of  the  city  is  located 
when  we  know  its  distance  and  direction  from  each  of  these 
lines. 

In  like  manner,  if  we  take  any  point  such  as  Pi  (Fig.  1)  in 
the  plane  of  the  paper,  its  position  is  fully  determined  when 
we  know  its  distance  and  direction  from  each  of  the  two  lines 
0  X  and  0  Y  which  intersect  each  other  at  right  angles  in 
that  plane.  This  method  of  locating  points  is  known  by  the 
name  of  The  Cartesian,  or  Bilijsteak  System.     The  lines  of 

1 


PLANE  ANALYTIC  GEOMETRY 


reference  0  X,  O  Y,  are  called  Co-ordinate  Axes,  and,  when 
read  separately,  are  distinguished  as  the  X-axis  and  the 
Y-AXIS.  The  point  0,  the  intersection  of  the  co-ordinate 
axes,  is  called  .the  Okigin  of  Co-ordinates,  or  simply  the 
Origin. 

The  lines   x'  and   y'  which   measure  the    distance  of   the 
point   Pi  from    the  Y-axis   and  the  X-axis   respectively,  are 


?z 

Y 

x' 

R 

0 

y 

Y 

P3 

P4 

Fig.  1. 

called  the  co-ordinates  of  the  point  —  the  distance  (x')  from 
the  Y-axis  being  called  the  abscissa  of  the  point,  and  the  dis- 
tance (?/')  from  the  X-axis  being  called  the  ordhiate  of  the 
point. 

2.  Referring  to  Fig.  1,  we  see  that  there  is  a  point  in  each 
of  the  four  angles  formed  by  the  axes  which  would  satisfy 
the  conditions  of  being  distantras'  from  the  Y-axis  and  distant 
y'  from  the  X-axis.  This  ambiguity  vanishes  when  we  com- 
bine the  idea  of  direction  with  these  distances.  In  the  case 
of  places  on  the  earth's  surface  this  difficulty  is  overcome  by 
using  the  terms  north,  south,  east,  and  ivest.  In  analytic  geome- 
try the  algebraic  symbols  -|-  and  —  are  used  to  serve  the  same 
purpose.     All  distances  measured  to  the  right  of  the  Y-axis 


CO-ORDINA  TES.  8 

are  called  jjositive  abscissas ;  those  measured  to  the  left, 
negative ;  all  distances  measured  above  the  X-axis  are  called 
positive  ordinates  ;  all  distances  beloiv,  7iegative.  With  this 
understanding,  the  co-ordinates  of  the  point  Pj  become  (x',  y')  ; 
of  P„  (-  x',  y')  ;  of  Pa,  (-  x',  -  y') ;  of  P,,  (x',    -  y'). 

3.  The  four  angles  which  the  co-ordinate  axes  make  with 
each  other  are  numbered  1,  2,  3,  4.  The  first  angle  is  above 
the  X-axis,  and  to  the  right  of  the  Y-axis ;  the  second  angle 
is  above  the  X-axis,  and  to  the  left  of  the  Y-axis ;  the  third 
angle  is  below  the  X-axis,  and  to  the  left  of  the  Y-axis ;  the 
fourth  angle  is  below  the  X-axis  and  to  the  right  of  the 
Y-axis. 

EXAMPLES. 

1.  Locate  the  following  points :        ■  • 

(-  1,  2),  (2,  3),  (3,  -  1),  (-  1,  -  1),  (-  2,  0),  (0,  1), 
(0,  0),  (3,  0),  (0,  -  4). 

2.  Locate  the  triangle,  the  co-ordinates  of  whose  vertices 
are, 

(0,  1),  (-  1,  -  2),  (3,  -  4). 

3.  Locate  the  quadrilateral,  the  co-ordinates  of  whose  ver- 
tices are, 

(2,  0),  (0,  3),  (-  4,  0),  (0,  -  3). 

What  are  the  lengths  of  its  sides  ? 

Ans.     VT3,  5,  5,  V 13. 

4.  The  ordinates  of  two  points  are  each  =  —  5  ;  how  is 
the  line  joining  them  situated  with  reference  to  the  X-axis  ? 

Ans.   Parallel,  below. 

5.  The  common  abscissa  of  two  points  is  a;  how  is  the 
line  joining  them   situated  ? 

6.  In  what  angles  are  the  abscissas  of  points  positive  ? 
In  what  negative  ? 

7.  In  what  angles  are  the  ordinates  of  points  negative  ? 
In  what  angles  positive  ? 


4  PLANE  ANALYTIC  GEOMETRY. 

8.  In  what  angles  do  the  co-ordinates  of  points  have  like 
signs  ?     In  what  angles  unlike  signs  ? 

9.  The  base  of  an  equilateral  triangle  coincides  with  the 
X-axis  and  its  vertex  is  on  the  Y-axis  at  the  distance  3  below 
the  origin ;  required  the  co-ordinates  of  its  vertices  ? 

Ans.     a  VT2,  0),  (0,  -  3),  (-x  Vl2,  0). 

10.  If  a  point  so  moves  that  the  ratio  of  its  abscissa  to  its 
ordinate  is  always  =  1,  what  kind  of  a  path  will  it  describe, 
and  how  is  it  situated  ? 

Ans.  A  straight  line  passing  through  the  origin,  and  mak- 
ing an  angle  of  45°  with  the  X-axis. 

11.  The  extremities  of  a  line  are  the  points  (2, 1),  (  —  1,  —  2) : 
construct  the  line. 

12.  If  the  ordinate  of  a  point  is  =  0,  on  which  of  the 
co-ordinate  axes  must  it  lie  ?     If  the  abscissa  is  =  0  ? 

13.  Construct  the  points  (—  2,  —  3),  (2,  3),  and  show  that 
the  line  joining  them  is  bisected  at  (0,  0). 

14.  Show  that  the  point .  (m,  n)  is  distant  Vm^  +  n"^  from 
the  origin. 

15.  Find  from  similar  triangles  the  co-ordinates  of  the 
middle  point  of  the  line  joining  (2,  4),  (1,  1). 

Ans.    (f,  f). 

THE   POLAR   SYSTEM. 

4.  Instead  of  locating  a  point  in  a  plane  by  referring  it  to 
two  intersecting  lines,  we  may  adopt  the  second  of  the  two 
methods  indicated  in  Art.  1.  The  point  Pi,  Fig.  2,  is  fully 
determined  when  we  know  its  distance  0  Pi  (=  ^)  and  direc- 
tion Pi  0  X  (=  0)  from  some  given  point  0  in  some  given 
line  0  X.  If  we  give  all  values  from  0  to  oo  to  r,  and  all 
values  from  0°  to  360°  to  $,  it  is  easily  seen  that  the  position 
of  every  point  in  a  plane  may  be  located. 

This  method  of  locating  a  point  is  called  the  Polar  System. 


CO-ORDINA  TES.  5 

The  point  0  is  called  the  Pole  ;  the  line  0  X,  the  Polar 
Axis,  or  Initial  Line;  the  distance  r,  the  Radius  Vector; 
the  angle  6,  the  Directional  or  Vectorial  Angle.  The 
distance  r  and  the  angle  6,  (r,  6),  are  called  the  Polar  Co- 
ordinates of  a  point. 

5.   In  measuring  angles  in  this  system,  it  is  agreed  (as  in 
trigonometry),  to  give  the  positive  sign  (-J-)  to  all  angles  meas- 


FlG.  2. 


ured  round  to  the  left  from  the  polar  axis,  and  the  opposite 
sign  (— )  to  those  measured  to  the  right.  The  radius  vector 
(r)  is  considered  as  positive  (-j-)  when  measured  from  the 
pole  toivard  the  extremity  of  the  arc  {&),  and  negative  (— ) 
when  measured  from  the  pole  away  from  the  extremity  of  the 
arc  (6).  A  few  examples  will  jnake  this  method  of  locating 
points  clear. 

If  r  =  2  inches  and  6  —  45°,  then  (2,  45°)  locates  a  point 
Pi  2  inches  from  the  pole,  and  on  a  line  making  an  angle 
of  -|-  45°  with  the  initial  line. 

If  r  =  —2  inches  and  0  =  45°,  then  (—  2,  45°)  locates  a 
point  Pg  2  inches  from  the  pole,  and  on  a  line  making  an 
angle  of  45°  with  the  initial  line  also ;  but  in  this  case  the 
point  is  on  that  portion  of  the  boundary  line  of  the  angle 
which  has  been  produced  backward  through  the  pole. 

If   r  =  2  inches  and  0  =  —  45°,  then    (2,  —  45°)  locates  a 


6  PLANE  ANALYTIC  GEOMETRY. 

point  P4  2  inches  from  the  pole,  and  out  on  a  line  lying  below 
the  initial  line,  and  making  an  angle  of  45'^  with  it. 

If  r=  -2  inches  and  0=  -  45°,  then  (  -  2,  -  45°  ) 
locates  a  point  Pg  directly  opposite  (with  respect  to  the  pole), 
the  point  P4,  (2,  -  45°). 

6.  While  the  usual  method  in  analytic  geometry  of  express- 
ing an  angle  is  in  degrees,  Tninictes,  and  seconds  (°,  ',  "),  it 
frequently  becomes  convenient  to  express  angles  in  terms  of 
the  angle  whose  arc  is  equal  in  length  to  the  I'adius  of  the 
measuring  circle.     This  angle  is  called  the  Circular  Unit. 

We  know  from  geometry  that  angles  at  the  centre  of  the 

same  circle  are  to  each  other  as  the  arcs  included  between 

their  sides ;   hence,   if   0  and  0'  be   two   central   angles,  we 

have, 

6  arc 

6'  ~  arc'  ' 

Let  6'  =  unit  angle ;  then  arc'  =  r  (radius  of  measuring 
circle). 

0  arc 

Hence  circular  unit         r 

,-.  r  6  =  arc  X  circular  unit. 
If  ^  =  360°,  common  measure,  then  arc  =  2  7r  r. 
Hence,  r  X  360°  =  2  tt  ?■  X  circular  unit. 

Therefore  the  equation, 

360°  =  2  TT  X  circular  unit,    ...   (1) 
expresses  the  relationship  between  the  two  units  of  measure. 

EXAMPLES. 

1.   What  is  the  value  in  circular  measure  of  an  angle  of  30°  ? 
Prom  (1)  Art.  6,  we  have, 

360°  =  30°  X  12  =  2  TT  circular  unit. 

.-.  30°  =  "  circular  unit. 
6 


CO-ORDINA  TES.  7 

2.  What  are  the  values  in  circular  measure  of  the  following 
angles  ? 

1°,  45°,  60°,  90°,  120°,  180°,  225°,  270°,  360°. 

3.  What  are  the  values  in  degrees  of  the  following:  angles? 


TT       TT 


3      3      5      17      5 


—  1     — >     —  TT,     —  TT,     —  TT,     —  TT,    —  TT,     —  TT,     — ■ 

3'  2    4    '8       4    '4    '8    '8    '6' 


9 


4.  What  is  the  unit  of  circular  measure  ? 

Ans.    57°,  17',  45". 

5.  Locate  the  following  points  : 

(2,  40°),  U,  |\  (-  4,  90),  (3,  -  135°),  (-  1,  -  180°), 

(  -  1,  -  ^),  {2,  -  fj,  (a,  ^^ 

6.  Locate  the  triangle  whose  vertices  are, 


2,  ^  i> 


3,^\ri,-'' 


7.  The  base  of  an  equilateral  triangle  (=  a)  coincides  with 
the  initial  line,  and  one  of  its  vertices  is  at  the  pole ;  re- 
quired the  polar  co-ordinates  of  the  other  two  vertices. 

.      Ans.     la,  Ij,   (^a,  0). 

8.  The   polar   co-ordinates    of    a  point    are  (2,  -  ].     Give 

three  other  ways   of   locating   the    same   point,   using  polar 
co-ordinates. 

-•  (-■¥)'(-'-¥)-^-V')- 

9.  Construct  the  line  the  co-ordinates  of  whose  extremities 


8  PLANE  ANALYTIC  GEOMETRY. 

10.  How  is  the  line,  the  co-ordinates  of  two  of  its  points 
being  I  3,-\  I  3,  '^  \  situated  with  reference  to  the  initial 
line  ?  Ans.    Parallel. 

Find  the  rectangular  co-ordinates  of  the  following  points  : 

11.  (s,  fj.  13.    (4, 

12.  (-3,  ^Y  14.   f-2,  |- 


LOCI. 


CHAPTEE   II. 
LOCI. 

7.  The  Locus  of  an  Equation  is  the  path  described  by  its 
generatrix  as  it  moves  in  obedience  to  the  law  expressed  in  the 
equation. 

The  Equation  of  a  Locus  is  the  algebraic  expression  of 
the  law  subject  to  which  the  generatrix  moves  in  describing  that 
locus. 

If  we  take  any  point  Pg,  equally  distant  from  the  X-axis 
and  the  Y-axis,  and  impose  the  condition  that  it  shall  so  move 


Fig.  3. 
that  the  ratio  of  its  ordinate  to  its  abscissa  shall  always  be 
equal  to  1,  it  will  evidently  describe  the  line    P3P1.      The 
algebraic  expression  of  this  law  is 

^  =  1,  or  2/  =  X, 

X 

and  is  called  the  Equation  of  the  Locus. 

The  line  P3P1  is  called  the  Locus  of  the  Equation.     Again  : 


10  PLANE  ANALYTIC  GEOMETRY. 

if  we  take  the  point  P4,  equally  distant  from  the  axes,  and 
make  it  so  move  that  the  ratio  of  its  ordinate  to  its  abscissa 
at  any  point  of  its  path  shall  be  equal  to  —  1,  it  will  describe 
the  line  P4  Po.     In  this  case  the  equation  of  the  locus  is 

^  =  —  1,  or  y  =  —  x, 

X 

and  the  line  P4  P2  is  the  locus  of  this  equation. 

8.  It  will  be  observed  in  either  of  the  above  cases  (the 
first,  for  example),  that  while  the  point  Pg  moves  over  the 
line  Pg  Pi,  its  ordinate  and  abscissa  while  always  equal  are 
yet  in  a  constant  state  of  change,  and  pass  through  all  values 
from  —  00,  through  0,  to  -\-  00.  For  this  reason  y  and  x  are 
called  the  Variable  or  General  Co-ordinates  of  the  line. 
If  we  consider  the  point  at  any  particular  position  in  its 
path,  as  at  P,  its  co-ordinates  {—  x',  —  y')  are  constant  in 
value,  and  correspond  to  this  position  of  the  point,  and  to 
this  position  alone.  The  variable  co-ordinates  are  represented 
by  X  and  y,  and  the  particular  co-ordinates  of  the  moving 
point  for  any  definite  position  of  its  path  by  these  letters 
with  a  dash  or  subscript ;  or  by  the  first  letters  of  the 
alphabet,  or  by  numbers.  Thus  {x',  y'),  {xi,  y^),  (a,  b),  (2,  2) 
correspond  to  some  particular  position  of  the  moving  point. 

EXAMPLES. 

1.  Express  in  language  the  law  of  which  ?/  =  3  cc  +  2  is  the 
algebraic  expression. 

Ans.  That  a  point  shall  so  move  in  a  plane  that  its  ordinate 
shall  always  be  equal  to  3  times  its  abscissa  plus  2. 

2.  A  point  so  moves  that  its  ordinate  -|-  a  quantity  a  is 
always  equal  to  ^  its  abscissa  —  a  quantity  h  ;  required  the 
alp'ebraic  expression  of  the  law. 

Ans.     y  -\-  a  ^  \x  —  h. 

3.  The  sum  of  the  squares  of  the  ordinate  and  abscissa  of 
a  moving  point  is  always  constant,  and  =  «,  ^ ;  what  is  the 
equation  of  its  path  ? 

Ans.     x^  -\-  y^  ^=  a^. 


LOCI.  11 

4.    Give  in  language  the  laws  of  which  the  following  are 
the  algebraic  expressions : 


•^               2 

X^  —  y^  =  —Q. 

3         2 

xy  =  16. 

if  =  -ix. 

4  ^2  -  5  ^2  =  -  18, 

2  cc2  -1-  3  2/'  = 

6. 

y' 

=  2  px. 

*V"  +  ^^^^  ==  '*^^^- 

9.  As  the  relationship  between  a  locus  and  its  equation 
constitutes  the  fundamental  conception  of  Analytic  Geometry, 
it  is  important  that  it  should  be  clearly  understood  before 
entering  upon  the  treatment  of  the  subject  proper.  We  have 
been  accustomed  in  algebra  to  treat  every  equation  of  the 
form  ?/  =  a;  as  indeterminate.  Here  we  have  found  that  this 
equation  admits  of  a  geometric  interpretation  ;  i.e.,  that  it  repre- 
sents a  straight  line  passing  through  the  origin  of  co-ordi- 
nates and  making  an  angle  of  45°  with  the  X-axis.  We  shall 
find,  as  we  proceed,  that  every  equation,  algebraic  or  transcen- 
dental, which  does  not  involve  more  than  three  variable  quan- 
tities, is  susceptible,  of  a  geometric  interpretation.  We  shall 
find,  conversely,  that  geometric  forms  can  be  expressed  alge- 
braically, and  that  all  the  properties  of  these  forms  may  be 
deduced  from  their  algebraic  equivalents. 

Let  us  now  assume  the  equations  of  several  loci,  and  let  us 
locate  and  discuss  the  geometric  forms  which  they  represent. 

10.  Locate  the  geomet7'ic  figure  tvhose  algebraic  equivalent  is 

y  =  ^x+2. 

We  know  that  the  point  where  this  locus  cuts  the  Y-axis  has 
its  abscissa  a;  =  0.  If,  therefore,  we  make  ic  =  0  in  the  equa- 
tion, we  shall  find  the  ordinate  of  this  point.  Making  the 
substitution  we  find  y  =  2.  Similarly,  the  point  where  the 
loous  cuts  the  X-axis  has  0  for  the  value  of  its  ordinate.    Mak- 


12 


PLANE  ANALYTIC  GEOMETRY. 


ing  2/  =  0  in  the  equation,  we  find  x  = 
the  axes  and  marking  on  them  the  points 


(0,  2), 


3 


0 


Drawing  now 


Now  make  x 


we  will  have  two  points  of  the  required  locus 
successively  equal  to 

1,  2,  3,  -1,  -2,  -  3,  etc. 

in  the  equation,  and  find  the  corresponding  values  of  y. 
convenience  let  us  tabulate  the  result  thus : 

Corresponding  Values  of  y 

5 


For 


Fig.  4. 


Locating  these  points  and  tracing  a  line  through  them  we 
have  the  required  locus.     This  locus  appears  to  be  a  straight 


LOCI.  13 

line  —  and  it  is,  as  we  shall  see  hereafter.  We  shall  see  also 
that  every  equation  of  the  first  degree  between  two  variables 
represents  some  straight  line.  The  distances  Oa  and  Ob 
which  the  line  cuts  off  on  the  co-ordinate  axes  are  called 
Intercepts.  In  locating  straight  lines  it  is  usually  sufficient 
to  determine  these  distances,  as  the  line  drawn  through  their 
extremities  will  be  the  locus  of  the  equation  from  which  their 
values  were  obtained. 

EXAMPLES. 

1.    Locate  the  geometric  equivalent  of 
1 


9 


y  —  X  ^  1  —  2  X. 


Solving  with  respect  to  y  in  order  to  simplify,  we  have, 

y=-2x  +  2. 
The  extremities  of  the  intercepts  are 

(0,2),     (1,0). 

Locating  these  points,  and  drawing  a  straight  line  through 
them,  we  have  the  required  locus. 

Construct  the  loci  of  the  following  equations  : 

2.    y  =  -2x  —  2.  1.    ''J-  +  2x  =  'Sx  —  y. 

Z.    y  =  ^x  —  1.  8.   2x  +  3y  =  7  —  y. 

4.    y  ^  ax  -\-  b.  9.    =  ^ . 

■^  2  3 

«1                     ^            ini2/-2,            2a; -2, 
0.    -  y  =  ex  —  a.  10.    1  —  -^ \-  X  ^ \-  y. 

2  2  3 

6.    2  ?/  =  3  a?.  11.    a?  —  2/  = y  —  2  a;. 

12.   Is   the   point    (2,    1)    on   the   line    whose    equation    is 
2/  =  2  a;  -  3  ?     Is  (6,  9)  ?     Is  (5,  4)  ?     Is  (0,  -  3)  ? 

Note.  —  If  a  point  is  on  a  line,  its  co-ordinates  must  satisfy 
the  equation  of  the  line. 


14  PLANE  ANALYTIC  GEOMETRY. 

13.  Which  of  the  following  points  are  on  the  locus  of  the 
equation  3  a?^  -f-  2  y'^  =  6  ? 

(2,  1),  (V2,  0),  (0,  V3),  (-  1,  3),  (-  V2,  0),  (2,  V3) 

14.  Write  six  points  which  are  on  the  line 

-  y  —  2  X  =  3y  —  6. 

15.  Construct  the  polygon,  the  equations  of  whose  sides  are 

y  ==  —  2  X  —  1,  y  =  x,  y  =  5. 

16.  Construct  the  lines  y  =  sx  -\-  b  and  y  =  sx  -{-  4,  and 
show  by  similar  triangles  that  they  are  parallel. 

11.    Discuss  and  construct  the  equation : 
x'^  -\-  y^  =  16. 

Solving  with  respect  to  y,  we  have, 
y  =±  Vl6  -  x^- 

The  double  sign  before  the  radical  shows  us  that  for  every 
value  we  assume  for  x  there  will  be  two  values  for  y,  equal 
and  with  contrary  signs.  This  is  equivalent  to  saying  that 
for  every  point  the  locus  has  above  the  X-axis  there  is  a  cor- 
responding point  below  that  axis.  Hence  the  locus  is  symmet- 
rical with  respect  to  the  X-axis.  Had  we  solved  the  equation 
with  respect  to  cc  a  similar  course  of  reasoning  would  have 
shown  us  that  the  locus  is  also  symmetrical  with  respect  to  the 
Y-axis.  Looking  under  the  radical  we  see  that  any  value  of  x 
less  than  4  (positive  or  negative)  will  always  give  two  real 
values  for  y  ;  that  x  =  -^4  will  give  y  =z  ^0,  and  that  any 
value  of  X  greater  than  J-  4  will  give  imaginary  values  for  y. 
Hence  the  locus  does  not  extend  to  the  right  of  the  Y-axis 
farther  than  cc  =  -j-  4,  nor  to  the  left  farther  than  x  =  —  4. 

Making  x  =  0,  we  have  ?/  =  J-  4 

"  2/  =  0,    "      "     X  =  i  4. 


LOCI. 


15 


Drawing  the  axes  and  constructing  the  points, 

(0,  4),  (0,  —  4),  (4,  0),  (—  4,  0),  we  have  four  points  of 
the  locus ;  i.e.,  B,  B',  A,  A'. 


Values  of  y 
+  3.8  and  —  3.8 
+  3.4  and  -  3.4 
+  2.6  and  -  2.6 
iO 

+  3.8  and  -  3.8 
+  3.4  and  —  3.4 
+  2.6  and  -  2.6 
±0 

Constructing  these  points  and  tracing  the  curve,  we  find  it 
to  be  a  circle. 

This  might  readily  have  been  inferred  from  the  form 
of  the  equation,  for  we  know  that  the  sum  of   the  squares 


ues  of  X 

C 

orresponding 

1 

a 

2 

u 

3 

a 

4 

(I 

-1 

a 

-2 

it 

-3 

ec 

-4 

u 

16  PLANE  ANALYTIC  GEOMETRY, 

of  the  abscissa  (OC)  and  ordinate  (CPi)  of  any  point  Pi 
in  the  circle  is  equal  to  the  square  of  the  radius  (OPi). 
We  might,  therefore,  have  constructed  the  locus  by  taking 
the  origin  as  centre,  and  describing  a  circle  with  4  as  a  radius. 

IS]"oTE.  ic  =  _J_  0  for  any  assumed  value  of  ^Z,  or  ?/  =  _[-  0, 
for  any  assumed  value  of  x  always  indicates  a  tangency.  Re- 
ferring to  the  figure  we  see  that  as  x  increases  the  values  of 
y  decrease  and  become  J-  0  when  x  =  4.  Drawing  the  line 
represented  by  the  equation  cf  =  4,  we  find  that  it  is  tangent 
to  the  curve.  We  shall  see  also  as  we  proceed  that  any  two 
coincident  values  of  either  variable  arising  from  an  assumed 
or  given  value  of  the  other  indicates  a  point  of  tangency. 

12.   Construct  and  discuss  the  equation 
9  »-2  +  16  2/2  =  144. 

Solving  with  respect  to  y,  we  have 


=W^ 


/144  -  9  £c2 


16 

a;  =  0  gives  ?/;=_[-  3 ; 
y  =  0      "      ic  .-=  ^  4. 
Drawing  the  axes  and  laying  off  these  distances,  we  have 
four  points  of  the  locus ;   i.e.,  B,  B',  A,  A'.     Fig.  6. 


Values  of  x 

Corresponding 

Values  of  y 

1 

4- 2.9  and  -  2.9 

2 

+  2.6     "    -2.6 

3 

_|_  2       "    2 

4 

±0 

-1 

+  2.9    "    -2.9 

-2 

4-2.6    "    -2.(^ 

-3 

+  2       "    -2 

-4 

±0 

Locating  these  points  and  tracing  the  curve  through  them, 
we  have  the  required  locus.  Referring  to  the  value  of  y  we 
see  from  the  double  sign  that  the  curve  is  symmetrical  with 
respect  to  the  X-axis.     The  form  of  the  equation  (containing 


LOCI. 


17 


only  the  second  powers  of  the  variables),  shows  that  the  locus 
is  also   symmetrical  with   respect   to   the   Y-axis.     Looking 


under  the  radical  we  see  that  any  value  of  x  between  the 
limits  4-  4  and  —  4  will  give  two  real  values  for  y  ;  and  that 
any  value  beyond  these  limits  will  give  imaginary  values  for 
y.      Hence  the  locus  is  entirely  included  between  these  limits. 

This  curve,  with  which  we  shall  have  more  to  do  hereafter, 
is  called  the  Ellipse. 

13.    Dismiss  and  construct  the  equation 

y^  =  4:  X. 

Solving,  we  have 

7/  =  J-  V4  X. 

We  see  that  the  locus  is  symmetrical  with  respect  to  the 
X-axis,  and  as  the  equation  contains  only  the  first  power  of 
X,  that  it  is  not  symmetrical  with  respect  to  the  Y-axis.  As 
every  positive  value  of  x  will  always  give  real  values  for  y, 
the  locus  must  extend  infinitely  in  the  direction  of  the  posi- 
tive abscissse ;  and  as  any  negative  value  of  x  will  render  y 


18 


PLANE  ANALYTIC   GEOMETRY. 


imaginary,  the  curve  can  have  no  point  to  the  left  of  the 
Y-axis.  Making  x  =  0,  we  find  y  =  J^Q-,  hence  the  curve 
passes  through  the  origin,  and  is  tangent  to  the  Y-axis. 
Making  y  =  0,  we  find  x  =0;  hence  the  curve  cuts  the 
X-axis  at  the  origin. 

Values  of  x  Corresponding 

1 

2 

3 

4 

From  these  data  we  easily  see  that  the  locus  of  the  equation 
is  represented  by  the  figure  below. 


Values  of  y 
+  2    and  -  2 
+  2.8  "    -  2.8 
+  3.4  "    -  3.4 

-f  4      "    —  4 


Fig.  7. 
This  curve  is  called  the  Parabola. 
14.    Discuss  and  construct  the  eqtcation 
4  a;2  _  9  2/2  =  36. 


Hence 


y 


=  W 


4  x^  -  36 


9 


We  see  from  the  form  of  the  equation  that  the  locus  must 
be  symmetrical  with  respect  to  both  axes.     Looking  under 


LOCI. 


19 


the  radical,  we  see  that  any  value  of  x  numerically  less  than 
+  3  or  —  3  will  render  y  imaginary.  Hence  there  is  no 
point  of  the  locus  within  these  limits.  We  see  also  that 
any  value  of  x  greater  than  +  3  or  —  3  will  always  give  real 
values  for  y.  The  locus  therefore  extends  infinitely  in  the 
direction  of  both  the  positive  and  negative  abscissae  from  the 
limits  a;  ^  -j-  3. 

Making  x  =  0,  we  find  ?/  =  -J^  2  V  —  1 ;  hence,  the  curve  ■ 
does  not  cut  the  Y-axis. 

Making   y  =  0,  we   find  a;  =  -J-  3 ;    hence,  the    curve   cuts 
the  X-axis  in  two  points  (3,  0),  (—  3,  0). 


Value  of  X. 

Corresponding. 

Values  of  y 

4 

+  1.7  and  -  1.7 

5 

+  2.6    "     -2.6 

6 

_|_3.4    "    _3.4 

-4 

+  1.7    "     -  1.7 

-5 

+  2.6    "     -2.6 

-6 

_i_3.4    "     -3.4 

Fig.  s. 


This  curve  is  called  the  Hyperbola. 


20  PLANE  ANALYTIC  GEOMETRY. 

15.  We  have  in  the  preceding  examples  confined  ourselves 
to  the  construction  of  the  loci  of  Rectangular  equations ; 
i.e.,  of  equations  whose  loci  were  referred  to  rectangular 
axes.     Let  us  now  assume  the  Polar  equation 

r  =  6  (1  -  cos  6) 

and  discuss  and  construct  it. 

Assuming  values  for  9,  we  find  their  cosines  from  some 
convenient  table  of  Natural  Cosines.  Substituting  these 
values,  we  find  the  corresponding  values  of  r. 


lues  of  9 

Values  of  cos  9 

Values  of  r 

0 

1. 

6  (1  -  1  )  =    0 

30° 

.86 

6  (1  -  .86)  =   .84 

60° 

.50 

6  (1  -  .50)  =  3. 

90° 

0 

6  (1  -  0  )  =  6. 

120° 

-.50 

6  (1  +  .50)  =  9. 

160° 

-.94 

6  (1  +  .94)  =  11.64 

180° 

-1. 

6  (1  +  1  )  =  12. 

200° 

-.94 

6  (1  +  .94)  =  11.64 

240° 

-.50 

6  (1  +  -50)  =  9. 

270° 

0 

6  (1  -  0  )  =  6. 

300° 

.50 

6  (1  -  50)  =  3. 

330°  .86  6  (1  -  .86)  =      .84 


Draw  the  initial  line  OX,  and  assume  any  point  0  as  the 
pole.  Through  this  point  draw  a  series  of  lines,  making  the 
assumed  angles  with  the  line  OX,  and  lay  off  on  them 
the  corresponding  values  of  r.  Through  these  points,  tra- 
cing a  smooth  curve,  we  have  the  required  locus. 


LOCI. 


21 


Fig.  9. 
This  curve,  from  its  heart-like  shape,  is  called  the  Cardioid. 

16.   Discuss  and  construct  the  transcendental  equation 
y  =  log  X. 

Note. — A  transcendental  equation  is  one  whose  degree 
transcends  the  power  of  analysis  to  express. 

Passing  to  equivalent  numbers  we  have  2^  =  x,  when  2  is 
the  base  of  the  system  of  logarithms  selected. 

As  the  base  of  a  system  of  logarithms  can  never  be  nega- 
tive, we  see  from  the  equation  that  no  negative  value  of  x  can 
satisfy  it.  Hence  the  locus  has  none  of  its  points  to  the  left 
of  the  Y-axis.  On  the  other  hand,  as  every  positive  value  of 
X  will  give  real  values  for  y,  we  see  that  the  curve  extends 
infinitely  in  the  direction  of  the  positive  abscissae. 

li  y  =  0,  then 

2°  =  aj  .•.  0  =  log  X  .•.  X  =  1. 


22 


PLANE  ANALYTIC  GEOMETRY. 


If  a;  =  0,  then 

2,y  =  0  .-.  y  =  log  0  .-.  2/  =  —  00. 

The  locus,  therefore,  cuts  the  X-axis  at  a  unit's  distance  on 
the  positive  side,  and  continually  approaches  the  Y-axis  with- 
out ever  meeting  it.  It  is  further  evident  that  whatever  be 
the  base  of  the  system  of  logarithms,  these  conditions  must 
hold  true  for  all  loci  whose  equations  are  of  the  form  ct"  =  x. 

Values  of  x  Corresponding  Values  of  y 

1  "  0 

2  "  1 
4  "  2 
8  "  3 
.5                            "                            -  1 

.25  "  —  2 

Locating  these  points,  the  curve  traced  through  them  will 
be  the  required  locus. 


Fig.  10. 


This  curve  is  called   the    Logarithmic    Curve,  its   name 
being  taken  from  its  equation. 


LOCI.  23 

17.  The  preceding  examples  explain  the  method  employed 
in  constructing  the  locus  of  any  equation.  While  it  is  true 
that  this  method  is  at  best  approximate,  yet  it  may  be  made 
sufficiently  accurate  for  all  practical  purposes  by  assuming 
for  one  of  the  variables  values  which  differ  from  each  other 
by  very  small  quantities.  It  frequently  happens  (as  in  the 
case  of  the  circle)  that  we  may  employ  other  methods  which 
are  entirely  accurate. 

18.  In  the  discussion  of  an  equation  the  first  step,  usually, 
is  to  solve  it  with  respect  to  one  of  the  variables  which  enter 
into  it.  The  question  of  which  variable  to  select  is  immate- 
rial in  principle,  yet  considerations  of  simplicity  and  conven- 
ience render  it  often  times  of  great  importance.  The  sole 
difficulty,  in  the  discussion  of  almost  all  the  higher  forms  of 
equations,  consists  in  resolving  them.  If  this  difficulty  can 
be  overcome,  there  will  be  no  trouble  in  tracing  the  locus  and 
discussing  it.  If,  as  frequently  happens,  no  trouble  arises  in 
the  solution  of  the  equation  with  respect  to  one  of  the  vari- 
ables, then  that  one  should  be  selected  as  the  dependent 
variable,  and  its  value  found  in  terms  of  the  other.  If  it  is 
equally  convenient  to  solve  the  equation  with  respect  to  either 
of  the  variables  which  enter  into  it,  then  that  one  should  be 
selected  whose  value  on  inspection  will  afford  the  simpler 
discussion. 

EXAMPLES. 

Construct  the  loci  of  the  following  equations : 

1.  2  2/  -  4  ^  +  1  =  0.  5.    ?/2  +  4  cc  =  0. 

2.  y^-x''  =  U.  6.    «2  +  2/'-25  =  0. 

3.  2  7/2  +  5x2  =  10.  7.    r^  =  a^coa2e. 

4.  4  a;2  —  9  ?/2  =  -  36.  8.   cc  =  log  y. 


24  PLANE  ANALYTIC  GEOMETRY. 

Construct  the  loci  of  the  following  : 

9.    a;2  -  2/2  =  0.  14.  a;2  —  a;  —  6  =  0. 

10.  x""  +  2  ax -\- a^  =  0.  15.  x^  +  cc  —  6  =  0. 

11.  ic2  -  ^2  ^  0.  16.  a:--^  +  4  a;  -  5  =  0. 

12.  2/2  _  9  _  0.  17.  x"'  -1x^12  =  0. 

13.  y''  —  2x7j  -[-  x^  =  0.  18.  a;2  +  7  £c  +  10  =  0. 

Note.  —  Factor  the  first  member :  equate  each  factor  to  0, 
and  construct  separately. 


THE  STRAIGHT  LINE. 


25 


CHAPTER   III. 

THE   STRAIGHT  LINE. 

19.  To  find  the  equation  of  a  straight  line,  given  the  angle 
which  the  line  makes  tvith  the  X-axis,  and  its  intercept  on  the 
Y-axis. 


Fig.  11. 

Let  C  S  be  the  line  whose  equation  we  wish  to  determine. 
Let  SAX  =  a  and  OB  =  b.  Take  any  point  P  on  the  line 
and  draw  PM  ||  to  OY  and  BN  ||  to  OX. 

Then  (OM,  MP)  =  (x,  ?/)  are  the  co-ordinates  of  P. 

From  the  figure  PM  =  PN  +  OB  =  BNtan  PBN  +  b,  but 
BN  =  OM  =  X,  and  tan  PBN  =  tan  SAX  =  tan  a. 

.*.  Substituting  and  letting  tan   «  =  s,  we  have, 


y  :=  sx  -\-  b 


(1) 


26  PLANE  ANALYTIC   GEOMETRY. 

Since  equation  (1)  is  true  for  any  point  of  the  line  SC,  it 
is  true  for  every  point  of  that  line ;  hence  it  is  the  equation  of 
the  line.  Equation  (1)  is  called  the  Slope  Equation  of  the 
Straight  Line  ;  s  (  =  tan  «)  is  called  the  slope. 

Corollary  1.   If  ^  =  0  in  (1),  we  have, 
y  =  sx  .  .  .   {2) 

for  the  slope  equation  of   a  line  which  passes  through  the 
origin. 

Cor.  2.   If  s  =  0  in  (1),  we  have 
y  =  h 

which  is,  as  it  ought  to  be,  the  equation  of  a  line  parallel  to 
the  X-axis. 

Cor.  3.  If  s  =  oo,  then «  =  90°,  and  the  line  becomes 
parallel  to  the  Y-axis. 

Let  the  student  show  by  an  independent  process  that  the 
equation  of  the  line  will  be  of  the  form  x  =  a. 

Scholium.  We  ha.ve  represented  by  «  the  angle  which  the 
line  makes  with  the  X-axis.  As  this  angle  may  be  either 
acute  or  obtuse,  s,  its  tangent,  may  be  either  positive  or  nega- 
tive. The  line  may  also  cut  the  Y-axis  either  above  or  beloiv 
the  origin;  hence,  b,  its  Y-intercept,  may  be  eilYiev positive  or 
negative.     Erom  these  considerations  it  appears  that 

y  =  —  sx  -{-  b 
represents  a  line  crossing  the  first  angle ; 

y  =  sx  -\-  b 
represents  a  line  crossing  the  second  angle ; 

y  =  —  sx  —  b 
represents  a  line  crossing  the  third  angle ; 

y  =  sx  —  b 
represents  a  line  crossing  the  fourth  angle. 


THE  STRAIGHT  LINE. 


27 


EXAMPLES. 

1.  The    equation   of   a  line    is  2  2/  +  a;  =  3 ;    required   its 
slope  and  intercepts. 

Solving  with  respect  to  y,  we  have, 

1       ,3 

1  3 

Comparing  with   (1)   Art.   19,  we  find  s  =  -  -  and  &  =  - 

=  Y-intercept.      Making    3/  =  0   in   the    equation,    we   have 
a;  =  3  =  X-intercept. 

2.  Construct  the  line  2  y  -\-  x  =  3. 

The  points  in  which  the  line  cuts  the  axes  are 


0,  ^  ] ,  and  (3,  0). 


Laying  these  points  off  on  the  axes,  and  tracing  a  straight 
line  through  them,  we  have  the  required  locus.  Or  otherwise 
thus :  solving  the  equation  with  respect  to  y,  we  have, 


y  = 


1  .3 

—  ^  +  ;r  • 

2  2 


Lay  off  OB  —  h  =  —;  draw 

BIST  II  OX  and  make  it  =  2,  also 
NP  II  OY  and  make  it  r=  +  1. 
The  line  through  P  and  B  is 
the  required  locus. 

p"Nr      1 

For    -^  =  i  =  tan  PEN 
NB        2 

^  —  tan  BAX. 


tan  BAX  =  s  =  —■ 


28 


PLANE  ANALYTIC   GEOMETRY. 


3.    Construct  the  line  2  y  —  x  ^  2>. 
Solving  with  respect  to  y,  we  have, 

1  ,   3 

2  2 

Lay  off  BO  =  ^»  =  - .     Draw  BN 

II  to   OX   and   make   it  =  2 ;    draw 
also  ISTP  II  to  OY  and  make  it  =  1. 
^    A  straight  line  through  P  and  B 
will  be  the  required  locus. 

PTsT       1 

For     —  =  -  =  tan  PBN  =  tan 
NB      2 

BAX  =  s.      Hence,  in  general,  BN  is  laid  off  to  the  right  or 

to  the  left  of  Y  according  as  the  coefficient  of  x  is  j)ositive 

or  negative. 

Give  the  slope  and  intercepts  of  each  of  the  following  lines 

and  construct : 

4.    22/  +  3cc-2  =  0. 


Ans.     s  =  —  —  ,  b  =  1,  <^  =  -^ 


5.   a;-22/+3  =  0. 


Ans.     s  =  —  ,  b  =  —  ,  a  =  —  3. 


6.   6aj  +  i-y  +  l  =  0. 


Ans.    s=—  12,  h  =  —2,a=—  —  . 

6 


7.    ^-'^  J^V- 


4. 


8. 


2/-1 


+  2x 


9.    a;+2  +  |- 


4. 


Note.  —  a  and  h  in  the  answers  above  denote  the  X-intercept 
and  the  Y-intercept,  respectively. 


THE  STRAIGHT  LINE.  29 

What  angle  does  each  of  the  following  lines  cross  ? 
10.  y  =  3  a;  +  1.  \2.    y  =  2x  —  l. 


11.    y  =  —X  -^2. 


13.   2/  =  -  3  cc  -  2. 


14.  Construct  the  figure  the  equation  of  whose  sides  are 
2y  +x-l  =  0,^y  =  2x  +  2,y=  -X  -1. 


15.    Construct  the  quadrilateral  the  equations  of  whose  sides 


are 


x  =  3,  y^— x-\-\,  y  =  2,  x  =  0. 


20.  To  find  the  equation  of  a  straight  line  in  terms  of  its 
intercepts. 


Fig.  12. 


Let  S  C  be  the  line. 

Then  OB  =b  =  Y-intercept,  and 
OA  =  a  =  X-intercept. 

The  slope  equation  of   a  line  we  have  determined  to  be 
Art.  19,  equation  (1), 

y  ^=  sx  -\-  b. 


30  PLANE  ANALYTIC  GEOMETRY. 

From  the  right  angled  triangle  AOB,  we  have, 

OR 
tan  GAB  =  —  tan  BAX  =  —  s  =  -— . 

OA 

^  _^ 

a 

Substituting  in  the  slope  equation,  we  have, 

y  = X  +b; 

a 

.-.   ^  +  ^  =  1   .   .   .    (1) 
a       0 

This  is  called  the  Symmetrical  Equation  of  the  straight 
line. 

Cor.  1.    It  a  -\-  and  b  -\-,  then  we  have, 

—  +  ^  =  1,  for  a  line  crossing  the  first  angle. 
a       b 

If  a  —  and  b  -\-,  then 

_^_|_2/_-j^-gg^  line  crossing  the  second  angle. 
a       b 

If  a  —  and  b  —,  then 

_^_y_^.g^  line  crossing  the  third  angle. 
a        b 

If  a  -f-  and  b  —,  then 

-  —  ^  =  1  is  a  line  crossing  the  fourth  angle. 
a        b 


EXAMPLES. 
1.    Construct  —  —  ^  =  1. 

Note.  —  Lay  off  3  units  on  the  X-axis  and  —  2  units  on  the 
Y-axis.     Join  their  extremities  by  a  straight  line. 

Across  which  angles  do  the  following  lines  pass  ? 


THE  STRAIGHT  LINE.  31 

Give  the  intercepts  of  each,  and  construct. 

2.   ^+^=1.  4.    -1-2^  =  1. 

3       2  2       4 

3.-^+^  =  1.  o.   ^-^=1. 

•3^3  57 

Write   the   slope   equations    of    the    following   lines,    and 
construct : 

6.   ^  _  ^  =  1. 
5       6 

Ans.    y  =  —  X  —  6. 
5 


7. 

X 

3 

7 

Ans.    y  =  — 7. 

o 

8. 

2  ^ 

^  =  -1. 
6 

A71S.    y  = X  —  2. 

^           3 

9. 

-y 

5 

^?ZS.     ?/  =    —  -X  —  1. 

5 

10.    Write  3/  =  sec  +  ^  in  a  symmetrical  form. 


Given  the  following  equations  of  straight  lines,  to  write 
their  slope  and  symmetrical  forms  : 

11.    2  2/  +  3  a;  -  7  =  a;  +  2.         13.    '^^^  =  3. 

X 

12  y ~ ^  —  ^  —  ^  14  ^  —  y  =  2x  —  1 

2  3    •  4  3     * 


32 


PLANE  ANALYTIC  GEOMETRY. 


21.  To  find  the  equation  of  a  straight  line  in  terms  of  the 
perpe7idicular  to  it  from  the  origin  and  the  directional  cosines 
of  the  'perpendicular. 

Note.  —  The  Directional  cosines  of  a  line  are  the  cosines  of 
the  angles  which  it  makes  with  the  co-ordinate  axes. 


Fig,  13. 


Let  CS  be  the  line. 

Let  OP  =  p,  BOP  =  7,  AOP  =  «. 

From  the  triangles  AOP  and  BOP,  we  have 


OA  =  OP  ,  OB  = -0£- 


that  is, 


cos  «  cos  y 


cos  a  COS  7 

Substituting  these  values  in  the  symmetrical  equation, 

Art.  20.  (1),    ?  -)-  ?^  =  1,  we  have,  after  reducing, 
ah 

X  cos  cc  -\-  1/  COS  Y  =:  p   .    .    .    (1) 

which  is  the  required  equation. 


THE  STRAIGHT  LINE.  33 

Since  y  =  90°  —  «,  cos  y  =  sin  a ;  hence 

X  cos  «  +  2/  sin  a  =  ^j  .   .  .   (2) 

This  form  is  more  frequently  met  with  than  that  given  in  (1) 
and  is  called  the  Normal  Equation  of  the  straight  line. 
CoR.  1.    If  a  =  0,  then 
X  =p 

and  the  line  becomes  parallel  to  the  Y-axis. 
CoR.  2.    If  a  =  90°,  then 

and  the  line  becomes  parallel  to  the  X-axis. 

22.  If  X  cos  «  +  2/  sin  a  =  p  be  the  equation  of  a  given  line, 
then  X  cos  «  +  ?/  sin  «  =^  ^^  _j_  (Z  is  the  equation  of  a  parallel 
line.  For  the  perpendiculars  p  and  j^  i  c?  coincide  in  direction 
since  they  have  the  same  directional  cosines ;  hence  the  lines 
to  which  they  are  perpendicular  are  parallel. 

Cor.  1.    Since 

p)  -^d  —  2^  ^  ^d 

it  is  evident  that  d  is  the  distance  between  the  lines.  If, 
therefore,  (x',  y')  be  a  point  on  the  line  whose  distance  from 
the  origin  is  j^  rt  d,  we  have 

x'  cos  «  +  y'  sin  a  =  p  ^  d. 

.-.  ^  d  =  x  cos  oi  -\-  y'  sin  «  —  p  .  .  .   (1) 

Hence  the  distance  of  a  point  {x,  y')  from  the  line 
ic  cos  "  +  ?/ sin  c£  =^  is  found  by  transposing  the  constant 
term  to  the  first  member,  and  substituting  for  x  and  y  the  co- 
ordinates x',  y'  of  the  point.  Let  us,  for  example,  find  the 
distance  of  the  point  {-^  3,  9)  from  the  line  x  cos  30°  -j-  y  sin 
30°  =  5. 

From  (1)         c^  =  V3  cos  30°  +  9  sin  30°  -  5 
=  V3.^|  +  9.|-5 

.-.  d  =  l. 


34  PLANE  ANALYTIC  GEOMETRY. 


From    Fig.    13   we  have    cos  «=-?-,  sin  a  =  -y  =  — ^^^^^^ 

a  b       Va^  +  ^2 

ab 

.-.  p  = 


Va'  +  b"" 
Hence  J^d  =  [  —  +  ^  -1   ' 


a         b  J  -yj  a^  -j-  b'^ ' 

is  the  expression  for  the  distance  of  the  point  (x',  y')  from  a 

line  whose  equation  is  of  the  form  ^  -f-  ?!'  =  1. 

a       b 

Let  the  student  show  that  the  expression  for  d  becomes 

VA2  +  B-^  ■ 

when  tlie  equation  of  the  line  is  given  in  its  general  form. 
See  Art.  24,  Equation  (1). 

EXAMPLES. 

1.  The  perpendicular  let  fall  from  the  origin  on  a  straight 
line  =  5  and  makes  an  angle  of  30°  with  X-axis  ;  required  the 
equation  of  the  line. 

Ans.     V3  X  -\-  y  =  10. 

2.  The  perpendicular  from  the  origin  on  a  straight  line 
makes  an  angle  of  45°  with  the  X-axis  and  its  length  =  ■\/2  5 
required  the  equation  of  the  line. 

Ans.     X  -\-  y  =  2. 

3.  What  is  the  distance  of  the  point  (2,  4)  from  the  line 

-  +  i^  =  1.  Ans.     d  =  -^  . 

4^2  VB" 

Find  the  distance  of  the  point  from  the  line  in  each  of  the 

following  cases  : 

4.  From  (2,  5)  to  ^  -1  =  1. 

5.  From  (3,  0)  to  -  -  ?^  =  1. 

^  '    ^        4       3 

6.  From  (0,  1)  to  2  ?/  —  a;  =  2. 

7.  From  (a,  c)  to  ?/  =  sx  +  b. 


THE   STRAIGHT  LINE.  35 

23.  To  find  the  equation  of  a  straight  line  referred  to 
oblique  axes,  given  the  angle  between  the  axes,  the  angle  lohich 
the  line  makes  with  the  X-axis  and  its  Y-intercept. 

Note.  —  Oblique  axes  are  those  whicli  intersect  at  oblique 
angles. 


Fig.  14. 

Let  CS  be  the  line  whose  equation  we  wish  to  determine, 
it  being  any  line  in  the  plane  YOX. 

Let  YOX  =  fi,  SAX  =  «,  OB  =  ^'. 

Take  any  point  P  on  the  line  and  draw 

PM  II  to  OY  and  ON  ||  to  SC ; 
then,  PM  =  2/,  OM  =  a;,  NOX  =  «,  NP  =  OB  =  Z>. 

From  the  figure 

2/  =  MN  +  NP  =  MN  +  ^>  .  .  .  (1) 

From  triangle  OISTM,  we  have, 

MN  ^  sin  NOM  . 
OM        sin  MNO  ' 

MN  sin « 


sin  {fS  —  a) 


Substituting  the  value  of  MN  drawn  from  this  equation  in 
(1),  we  have, 


sm  « 


Sill  «  I      ,  /o\ 

y  =  ^-^^ r  ^  +  ^    •  ■    •  (2) 

sm  ((3  —  u) 


36  PLANE  ANALYTIC  GEOMETRY. 

This  equation  expresses  the  relationship  between  the  co- 
ordinates of  at  least  one  point  on  the  line.  But  as  the  point 
selected  was  any  point,  the  above  relation  holds  good  for 
every  point,  and  is,  therefore,  the  algebraic  expression  of  the 
law  which  governed  the  motion  of  the  moving  point  in  de- 
scribing the  line.     It  is  therefore  the  equation  of  the  line. 

CoK.  1.    If  ^»  =  0,  then 

y==-^^x...   (3) 
sm  (/3  —  «) 

is  the  general  equation   of   a  line  referred  to  oblique  axes 
passing  through  the  origin. 

Cor.  2.    If  5  =  0  and  «  =  0,  then 
y  =  0  .  .  .   (4) 

the  equation  of  the  X-axis. 

CoR.  3.    If  &  =  0  and  /?  =  «,  then 
a;  =  0  ...  (5) 
the  equation  of  the  Y-axis. 

CoR.  4.    If  /3  =  90° ;  i.e.,  if  the  axes  are  made  rectangular, 

then 

y  =  tan  ft  X  -\-  b. 

But  tan  <t  =  s  .-.  y  =  sx  -\-  b. 

This  is  the  slope  equation  heretofore  deduced.  See  Art. 
19  (1). 

Cor.  5.   If  )8  =  90°  and  b  =  0,  then 

y  =  sx.     See  Art.  19,  Cor.  1. 

EXAMPLES. 

1.  Find  the  equation  of  the  straight  line  which  makes  an 

angle  of  30°  with  the  X-axis  and  cuts  the  Y-axis  two  units 

distant  from  the  origin,  the   axes  making  an   angle  of   60° 

with  each  other. 

Ans.    y  =  X  -\-  2. 

2.  If  the  axes  had  been  assumed  rectangular  in  the  exam- 
ple above,  what  would  have  been  the  equation  ? 

Ans.     y  =  — — -  +  2. 
V3 


THE  STRAIGHT  LINE.  37 

3.  The  co-ordinate  axes  are  inclined  to  each  other  at  an 
angle  of  30°,  and  a  line  passing  through  the  origin  is  inclined 
to  the  X-axis  at  an  angle  of  120°,  required  the  equation  of  the 
line. 

Ans.     2/  =  -  I  V3  • 

24.  Every  equation  of  the  first  degree  between  two  variables 
is  the  equation  of  a  straight  liiie. 

Every  equation  of  the  first  degree  between  two  variables 
can  be  placed  under  the  form 

Aa;  +  By  +  C  =  0  .  .  .  (1) 

in  which  A,  B,  and  C  may  be  either  finite  or  zero. 

Suppose  A,  B,  and  C  are  not  zero.  Solving  with  respect  to 
2/,  we  have, 

V  =  -  —  ^  -  —  ...   (2) 
■^  B  B  ^  ^ 

Comparing  equation  (2)  with  (1)  Art.  19,  we  see  that  it 
is   the   equation   of  a   straight   line   whose   Y-intercept  b  = 

and  whose  slope  s  = ;    hence    (1),   the    equation 

B  B 

from  which  it  was  derived  is  the  equation  of  a  straight  line. 

If  A  =  0,  then   y  =  -  -^ , 
B 

the  equation  of  a  line  parallel  to  the  X-axis. 

If  B  =  0,  then  x  =  -—, 
A  ' 

the  equation  of  a  line  parallel  to  the  Y-axis. 
If  C  =  0,  then   y  = ■  x, 

the  equation  of  a  line  passing  through  the  origin. 

Hence,  for  all  values  of  A,  B,  C  equation  (1)  is  the  equa- 
tion of  a  straight  line. 


88  PLANE  ANALYTIC  GEOMETRY. 

25.  To  find  the  eqiiation  of  a  straight  line  ^jassing  through 
a  given  ^oint. 

Let  {x',  y')  be  the  given  point. 

Since  the  line  is  to  be  straight,  its  equation  must  be 

y  =  sx-\-h  .  .  .   (1) 

in  which  s  and  h  are  to  be  determined. 

Now,  the  equation  of  a  line  expresses  the  relationship  which 
exists  between  the  co-ordinates  of  every  point  on  it ;  hence 
its  equation  must  be  satisfied  when  the  co-ordinates  of  any 
point  on  it  are  substituted  for  the  general  co-ordinates  x  and 
y.     We  have,  therefore,  the  equation  of  condition. 

y'  =  sx'-{-b  ...  (2) 

But  a  straight  line  cannot  in  general  be  made  to  pass 
through  a  given  point  (x',  y'),  cut  off  a  given  distance  (h)  on 
the  Y-axis,  and  make  a  given  angle  (tan.  =  s)  with  the  X-axis. 
We  must  therefore  eliminate  one  of  these  requirements.  By 
subtracting  (2)  from  (1),  we  have, 

y  -  y'  =  s  {x  -  x')   .   .   .    (3) 

which  is  the  required  equation. 
CoR.  1.    If  x'  =  0,  then 

y  -y'  =sx  .  .  .   (4:) 

is  the  equation  of   a  line  passing  through  a  point  on   the 
Y-axis. 

Cor.  2.    If  t/  =  0,  then 

y  =s  (x  -x')  .  .  .  (5) 

is  the  equation  of  a  line  passing  through  a  point  on  the  X-axis. 

CoR.  3.    If  x'  =  0  and  i/  =  0,  then 
y  ^  sx 

is   the  equation   (heretofore   determined),  of  a  line  passing 
throuarh  the  origin. 


THE   STRAIGHT  LINE.  39 

EXAMPLES. 

1.  Write  the  equation  of  several  lines  which  pass  through 
the  point  (2,  3). 

2.  What  is  the  equation  of  the  line  which  passes  through 

(1,-2),  and   makes   an   angle  whose   tangent  is  2  with  the 

X-axis  ? 

Ans.  ?/  =  2  a;  —  4. 

3.  A  straight  line  passes  through  (—  1,  —  3),  and  makes  an 
angle  of  45°  with  the  X-axis.     What  is  its  equation  ? 

Ans.   y  =  X  —  2. 

4.  Required  the  equations  of  the  two  lines  which  contain 
the  point  {a,  h),  and  make  angles  of  30°  and  60°,  respectively 
with  X-axis. 

^715.     y  -  b  =  ^  ~  ^  ;  y  —  b  =  ^^  .  (x  —  a). 

V3 

26.  To  find  the  equation  of  a  straight  line  passing  through 
two  given  points. 

Let  (cc',  y'),  (x",  y")  be  the  given  points. 
Since  the  line  is  straight  its  equation  must  be 

y  =  sx  +b  .  .  .   (1) 

in  which  s  and  b  are  to  be  determined. 

Since  the  line  is  required  to  pass  through  the  points  (x',  y'), 
(x",  y"),  we  have  the  equations  of  condition. 

y'  =  sx'  ^b  .   .   .   (2) 
y"  =  sx"  +  ^»    .    .    .    (3) 

As  a  straight  line  cannot,  in  general,  be  made  to  fulfil  more 
than  two  conditions,  we  must  eliminate  two  of  the  four  con- 
ditions expressed  in  the  three  equations  above. 

Subtracting  (2)  from  (1),  and  then  (3)  from  (2),  we  have,    . 

y  —  y'  =  s  {x  —  x') 
y'  —  y"  ^  s  (x'  —  x") 


40  PLANE  ANALYTIC  GEOMETRY. 

Dividing  these,  member  by  member,  we  have, 

y  —  y'   X  —  x' 

y'  -  y"  ~  x'  -  x"  ' 

Hence  y  —  y'  =  ^^  ~  ^^^  (x  —  x')  .  .  .  (4) 

is  the  required  equation. 
Cor.  1.    If  y'  =  y",  then 

y  —  ij  =  0,ovy  =  y', 

which  is,  as  it  should  be,  the  equation  of  a  line  ||  to  the  X-axis. 
CoK.  2.    If  x'  =  x",  then 

X  —  ic'  =  0,  or  cc  =  ic', 

which  is  the  equation  of  a  line  ||  to  the  Y-axis. 


EXAMPLES. 

1.  Given  the  two  points  (—  1,  6),  (—  2,  8) ;  required  both 
the  slope  and  symmetrical  equation  of  the  line  passing  through 
them. 

Ans.     2/=-2a;+4,  1  +  ^  =  1. 

2.  The  vertices  of  a  triangle  are  (—  2,  1),  (—  3,-4)  (2,  0)  ; 
required  the  equations  of  its  sides. 

r  y  =  5  x  +  11 
Ans.      •<4x  —  5?/  =  8 
(47/  +  *  =  2. 

Write  the  equations  of  the  lines  passing  through  the  points  : 

3.  (-2,  3),  (-3,-1)  6.    (5,2),  (-2,4) 

Ans.    t/  =  4  cc  +  11.  Ans.    7  y  -\- 2  x  =  24:. 

4.  (1,4),  (0,0)  7.    (2,0),  (-3,0) 

Ans.   y  =  4:X.  Ans.  y  =  0. 

5.  (0,2),  (3,-1)  8.    (- 1 -3),  (- 2,  4) 

Ans.    y  -\-  X  =  2.  Ans.    ?/  +  7  a;  -f- 10  =  0. 


THE  STRAIGHT  LINE. 


41 


27.    To  find  the  length  of  a  line  joining  two  given  points. 


Fig.  15. 

Let  (x',  y'),  (x",  y")  be  the  co-ordinates  of  the  given  points 
P',  P".     L  =  PT"  =  required  length. 

Draw  P"B  and  P'A  ||  to  OY,  and  P'C  ||  to  OX. 
We  see  from  the  figure  that  L  is  the  hypothenuse  of  a  right 
angled  triangle  whose  sides  are 

P'C  =  AB  =  OB  -  OA  =  x"  -  x',  and 
P"C  =  P"B  -  BC  =  y"  -  y'. 
Hence, 

PT"  =  L  =  -^{x"  -  x'Y  +  {y"  -  y'Y  ...   (1) 

CoR.  1.   If  x'  =  0  and  y'  =  0,  the  point  P'  coincides  with 
the  origin,  and  we  have 

L  =  ^x"^-\-y"^  •  •  •  (2) 
for  the  distance  of  a  point  from  the  origin. 

EXAMPLES. 
1.    Given  the  points  (2,  0),  (—  2,  3)  ;  required  the  distance 
between  them ;  also  the  equation  of  the  line  passing  through 
them. 

A71S.    1j  =  5,  4  y  -\-  3  x  ^  6. 


42  PLANE  ANALYTIC  GEOMETRY. 

2.  The  vertices  of  a  triangle  are  (2,  1)  (-  1,  2)  (  -  3,  0) ; 
what  are  the  lengths  of  its  sides  ?  _ 

Ans.    V8,  Vio^  V2a 

Give  the  distances  between  the  following  points : 

3.  (2,  3),  (1,  0)     _  7.    (-  3,  2),  (0,  1) 

Ans.     VlO. 

4.  (4,  -  5),  (6,  -_1)  8.    (-  2,-1),  (2,  0) 

Ans.     V20. 

5.  (0,  2),  (-  1,  0)_  9.    {a,  b),  (c,  d) 

Ans.     V5. 

6.  (0,  0),  (2,  0)  10.    (-  2,  3),  (-  a,  b). 

Ans.  2. 

11.  What  is  the  expression  for  the  area  of  a  triangle  whose 
vertices  are         {x',  y'),  {x",  y"),  {x"',  ij'")  ? 

Ans.    Area  =  |  [x'  (y"-  y'")  +  x"  (tj'" -  y')  +  x'"  {y' -  y")^ 

28.  To  find  the  intersection  of  two  lines  given  by  their 
equations. 

Let  y  =  sx  -\-  b,  and 

y  =  s^x  -\-  b' 

be  the  equations  of  the  given  lines. 

Since  each  of  these  equations  is  satisfied  for  the  co-ordinates 
of  every  point  on  the  locus  it  represents,  they  must  at  the 
same  time  be  satisfied  for  the  co-ordinates  of  their  point  of 
intersection,  as  this  point  is  common  to  both.  Hence,  for  the 
co-ordinates  of  this  point  the  equations  are  simultaneous.  So 
treating  them,  we  find 

b  -h'         .  s'b  -  sb' 

-,  and  y 


s  —  s  s  —  s 

for  the  co-ordinates  of  the  required  point, 

EXAMPLES. 

1.    Find  the  intersection  of  y  =  2  x  -\-  1  and  2  ?/  =  a;  —  4. 

Ans.    (-2,-3). 


THE   STRAIGHT  LINE.  43 

2.  The  equations  of  the  sides  of  a  triangle  are 

2y  =  3x-{-l,y-\-x  =  l,2y  +  4.x  =  -3; 
required  the  co-ordinates  of  its  vertices. 

3.  Write  the  equation  of  the  line  which  shall  pass  through 
the  intersection  of  2  y  -\- Zx  -\- 2  =  0  and  3t/  —  cc  —  8=0, 
and  make  an  angle  with  the  X-axis  whose  tangent  is  4. 

Ans.    y  =  4:X-\-10. 

4.  What  are  the  equations  of  the  diagonals  of  the  quadri- 
lateral the  equations  of  whose  sides  are  y  —  a:;  -f- 1  =  0, 
^y  =  -x  +  2,  y  =  3x-f2,  andy  +  2a;-4-2  =  0? 

Ans.     23?/  —  9a;+2  =  0,  32/-30a;  =  6. 

5.  The  equation  of  a  chord  of  the  circle  whose  equation  is 
x'^  ^  y'^  =^  10  is,  y  =  X  -\- 2 ;  required  the  length  of  the  chord. 

Ans.     L  =  V32y 

29.   If  Ax  +  By  +  C  =  0  .  .  .  (1) 

and  A'a;  -f  B't/  +  C  =  0  .  .  .   (2) 

be  the  equations  of  two  straight  lines,  then 

Aic  +  By  +  C  4-  II  {Mx  +  B't/  +C')  =  0  ...  (3) 

(K  being  any  constant  quantity)  is  the  equation  of  a  straight 
line  which  passes  through  the  intersection  of  the  lines  repre- 
sented by  (1)  and  (2).  It  is  the  equation  of  a  straight  line 
because  it  is  an  equation  of  the  first  degree  between  two 
variables.  See  Art.  24.  It  is  also  the  equation  of  a  straight 
line  which  passes  through  the  intersection  of  (1)  and  (2), 
since  it  is  obviously  satisfied  for  the  values  of  x  and  y  which 
simultaneously  satisfy  (1)  and  (2). 

Let  us  apply  this  principle  to  find  the  equation  of  the  line 
which  contains  the  point  (2,  3)  and  which  passes  through  the 
intersection  oi  y  =  2x  -\-l  and  2  y  -\-  x  —  2. 


44 


PLANE  ANALYTIC   GEOMETRY. 


From  (3)  we  have  y— 2x  — l-\-^{2y-\-x  — 2)=0 
for  the  equation  of  a  line  which  passes  through  the  intersec- 
tion of  the  given  lines.     But  by  hypotheses  the  point  (2,  3)  is 
on  this  line  ;  hence  3  —  4  —  l+K(6  +  2  —  2)=0 

.•.K  =  l. 
3 

Substituting  this  value  for  K  we  have, 

y-2x-l+^(2y  +  x-2),  =  0 
or,  y  —  X  —  1  =  0 

for  the  required  equation.  Let  the  student  verify  this  result 
by  finding  the  intersection  of  the  two  lines  and  then  finding 
the  equation  of  the  line  passing  through  the  two  points. 

30.    To  find   the   angle  between    two    lines   given   by   th&ir 
equations. 

M, 


Fig.  16. 


Let 


y  =^  sx-\-b,  and 
y  z=  s'x-\-  V 
be  the  equations  of  SC  and  MIST,  respectively ;  then 
s  =  tan  a  and  s'  =  tan  «'. 


THE  STRAIGHT  LINE.  45 


From  the  figures 

a'  =  cp  -[-  a 
.*.  (p  =  a'  —  «. 

From  trigonometry, 


,  i.       /  /        \         tan  a'  —  tan  a 

tan  fp  =  tan  (a  —  w)  = . 

1  -|-  tan  «  tan  « 

,-.  substituting 

*"° "  =  frt"'  ■  •  •  (^^• 

Or,  <p=tai,-'-5^  .  .  .  (2). 

1  +  ss 

Cor.  1.   If  s  ^  s',  then 

<jD  =  tan  ~^  0  .•.  (jD  =  0. 

.-.  the  lines  are  parallel. 
Cob.  2.    If  1  +  ss'  =  0,  then 

<p  =  tan  ~^  00  .-.  (jD  =  90° 
.*.  the  lines  are  perpendicular. 

ScHOL.     These  results  may  be  obtained  geometrically. 
If  the  lines  are  parallel,  then.  Fig.  16, 

If  they  are  perpendicular 
«'  =  90°  +  « 

.-.  tan  a'  =  s'  =  tan    (90°  +  «)  =  —  cot  «  = ^  . 

tan  a 

.:  1  -}-  ss'  =  1  -\-  tan  a  tan  «'  =  1  -|-  tan  «  ' 


tan  a 
1-1=0. 

EXAMPLES. 

1.   What  is  the  angle  formed  by  the  lines  y  —  x  —  1  ^  0 
and2?/  +  2x  +  l  =  0? 

Ans.    <p  =  90°. 


46  PLANE  ANALYTIC  GEOMETRY. 

2.  Required  the  angle  formed  by  the  lines  2/-|-3a:;  —  2  =  0 

and  2  3/  -f  6  a;  +  8  =  0. 

Ans.    cp  =  0. 

3.  Eequired  the  equation  of  the  line  which  passes  through 
(2,  —  1)  and  is 

(a).  Parallel  to2?/-3a^  —  5  =  0. 

(b)  Perpendicular  to2y  —  3x  —  5=0. 

Ans.  (a)  Sx  —  2y  =  S,     (b)  3y-^2x  =  l. 

4.  Given  the  equations  of  the  sides  of  a  triangle 

7/  =  2x-\-l,  y=  —  X  -\-2  and  y  =  —  3;  required. 

(a)  The  angles  of  the  triangle. 

(b)  The  equations  of  the  perpendiculars  from  vertices  to 
sides. 

(e)    The  lengths  of  the  perpendiculars. 

5.  What  relation  exists  between  the  following  lines : 

y  ^  sx  -\-  b. 
y  :=  sx  —  3. 
y  =z  sx  -\-  6. 
y  ^  sx  -\-m. 

6.  What  relation  exists  between  the  following: 

y  =  sx  -{-b. 
y  =  —  sx  -\-  c. 

7.  Pind  the  co-ordinates  of  the  point  in  which  a  perpen- 
dicular through  (—2,  3)  intersects  y  —  2x-\-l  =  0. 

Ans.     (^     1 
\5'  5 

8.  Pind  the  length  of  the  perpendicular  let  fall  from  the 
origin  on  the  line  2  y  -\-x  =  4. 

Ans.     L  =  -  VSa 

9.  If  Ax-\-By-\-G=  0,  A'x  +  B'y  -f  C  =  0,  and  A"x  + 
Wy-\-C"  =  0  be  the  equations  of  three  straight  lines,  and  I, 
m,   and   n   be    three    constants   which    render   the    equation 


THE  STRAIGHT  LINE.  47 

Z  (Ax  +  By  -f-  C)  -f-  m  (A'a;  +  B'y  +  C)  +  7i  {M'x  +  B"?/  + 
C")  =  0  an  identity,  then  the  tliree  lines  meet  in  a  point. 

10.  Find  the  equation  of  the  bisector  of  the  angle  between 
the  two  lines  Aa;  +  By  +  C  =  0  and  K'x  +  B'tj  +  C  =  0. 

Ans      Ax'  +  By  +  C  _  ^  (A^x  +  B^y  +  CQ 

Va^  +  b^  Va'2  +  b'2 

GENERAL   EXAMPLES. 

1.  A  straight  line  makes  an  angle  of  45°  with  the  X-axis 
and  cuts  off  a  distance  =  2  on  the  Y-axis ;  what  is  its  equation 
when  the  axes  are  inclined  to  each  other  at  an  angle  of  75°  ? 

Ans.     y  =  ^2  x  -\-2. 

2.  Prove  that  the  lines  y  =  x  -\-  1,  y  =  2x+2  and 
y  =  3  a;  -f-  3  intersect  in  the  point  (—  1,  0). 

3.  If   (x',  y')  and  (x",  y")  are  the  co-ordinates  of  the  ex- 

.  .  [  x"  -K- x'     v"  A- i/\ 

tremities  of  a  line,  show  that      — ^ — ,  -^ — ni^  ]  are  the  co- 

V      2       '        2       y 

ordinates  of  its  middle  point. 

4.  The  equations  of  the  sides  of  a  triangle  are  y  =  x  -\-l, 
x  =  4z,  y  =  —  X  —  1;  required  the  equations  of  the  sides  of 
the  triangle  formed  by  joining  the  middle  points  of  the  sides 
of  the  given  triangle. 

ry  =  —  x  +  4 
Ans.      J  y  =  x  —  4 
(  2  X  =  3. 

5.  Prove  that  the  perpendiculars  erected  at  the  middle 
points  of  the  sides  of  a  triangle  meet  in  a  common  point. 

Note.  —  Take  the  origin  at  one  of  the  vertices  and  make 
the  X-axis  coincide  with  one  of  the  sides.  Find  the  equations 
of  the  sides  ;  and  then  find  the  equations  of  the  perpendiculars 
at  the  middle  points  of  the  sides.  The  point  of  intersection 
of  any  two  of  these  perpendiculars  ought  to  satisfy  the  equa- 
tion of  the  third. 


48  PLANE  ANALYTIC  GEOMETRY. 

6.  Prove  that  the  perpendiculars  from  the  vertices  of  a 
triangle  to  the  sides  opposite  meet  in  a  point. 

7.  Prove  that  the  line  joining  the  middle  points  of  two  of 
the  sides  of  a  triangle  is  parallel  to  the  third  side  and  is  equal 
to  one-half  of  it. 

8.  The  co-ordinates  of  two  of  the  opposite  vertices  of  a 
square  are  (2,  1)  and  (4,  3)  ;  required  the  co-ordinates  of  the 
other  two  vertices  and  the  equations  of  the  sides. 

Ans.    (4,  1),  (2,^)-y  =  l,y  =  3,x  =  2,x  =  4. 

9.  Prove  that  the  diagonals  of  a  parallelogram  bisect  each 
other. 

10.  Prove  that  the  diagonals  of  a  rhombus  bisect  each  other 
at  right  angles. 

11.  Prove  that  the  diagonals  of  a  rectangle  are  equal. 

12.  Prove  that  the  diagonals  of  a  square  are  equal  and  bi- 
sect each  other  at  right  angles. 

13.  The  distance  between  the  points  {x,  y)  and  (1,  2)  is  =  4 ; 
give  the  algebraic  expression  of  the  fact. 

Ans.  (x  -  ly  +  (2/  -  2)2  =  41 

14.  The  points  (1,  2),  (2,  3)  are  equi-distant  from  the  point 
(x,  y).     Express  the  fact  algebraically. 

{X  -  ly  -f  (7/  -  2)2  =  {x-  2)2  +  (y  -  3)2;  or,  X  +  2/  =  4. 

15.  A  circle  circumscribes  the  triangle  whose  vertices  are 
(3,  4),  (1,  —  2),  (—  1,  2)  ;  required  the  co-ordinates  of  its  centre. 

Ans.    (2,  1). 

16.  What  is  the  expression  for  the  distance  between  the 
points  (x",  y"),  (x',  y'),  the  co-ordinate  axes  being  inclined 
at  an  angle  /?  ? 

Ans.   L  =  V(cc"  -  x'Y  +  {y"  -  yj  +  2  (x"  -  x!)  {y"  -  y')  cos  ^. 


THE  STRAIGHT  LINE.  49 

17.  Given  the  perpendicular  distance  (^j)  of  a  straight  line 
from  the  origin  and  the  angle  (")  Avhich  the  perpendicular 
makes  with  the  X-axis  ;  required  the  polar  equation  of  the  line. 


Ans. 


V 


cos  {Q  —  «)  ■ 

18.  Required  the  length  of  the  perpendicular  from  the 
origin  on  the  line  J  +  |  =  1.  ^^^_    2.4 

19.  What  is  the  equation  of  the  line  which  passes  through 
the  point  (1,  2),  and  makes  an  angle  of  45°  with  the  line 
whose  equation  isy  +  2cc  =  l? 

Ans.     y^-l.+l. 
^  3  3 

20.  One  of  two  lines  passes  through  the  points  (1,  2), 
(—  4,  —  3),  the  other  passes  through  the  point  (1,  —  3),  and 
makes  an  angle  of  45°  with  the  first  line ;  required  the 
equations  of  the  lines. 

Ans.    y  =  X  -{-  1,  and  ?/  =  —  3,  or  cc  =  1. 

21.  If  />  =  0  in  the  normal  equation  of  a  line,  through 
what  point  does  the  line  pass,  and  what  does  its  equation 
become  ?  A^is.    (0,  0)  ;  y  ^  s  x. 

22.  Required  the  perpendicular  distance  of  the  point  (r  cos  0, 
r  sin  &),  from  the  line  x  cos  6  -\-  y  sin  6  =2^-  Ans.     r  —  ^j. 

23.  Given  the  base  of  a  triangle  =  2  a,  and  the  difference 
of  the  squares  of  its  sides  =  4  c^  Show  that  the  locus  of 
the  vertex  is  a  straight  line. 

24.  What  are  the  equations  of  the  lines  which  pass  through 
the  origin,  and  divide  the  line  joining  the  points  (0,  1),  (1,  0), 
into  three  equal  parts.  Ans.    2x  =  y,2  y  ^  x. 

25.  If  (x',  y')  and  {x",  y")  be  the  co-ordinates  of  two  points, 

show  that  the  point  (  — '■ — ^^^ — ,  — ^^ — ~ — ^  Vlivides  the  line 
\     m  -\-  n  in  -\-  n    J 

joining  them  into  two   parts  which  bear  to   each  the  ratio 

m  :  n. 


50 


PLANE  ANALYTIC  GEOMETRY. 


CHAPTEE   IV. 


TRANSFORMATION    OF   CO-ORDINATES. 


31.  It  frequently  happens  that  the  discussion  of  an  equa- 
tion and  the  deduction  of  the  properties  of  the  locus  it 
represents  are  greatly  simplified  by  changing  the  position  of 
the  axes  to  which  the  locus  is  referred,  thus  simplifying  the 
equation,  or  reducing  it  to  some  desired  form.  The  operation 
by  which  this  is  accomplished  is  termed  the  Transformation 

OF    CO-ORDINATES. 


Fig.  17. 

The  equation  of  the  line  PC,  Fig.  17,  is 

y  =  SX  -{■  b 

when  referred  to  the  axes  Y  and  X.     If  we  refer  it  to  the 
axes  Y  and  X'  its  equation  takes  the  simpler  form 

y  :=  sx'. 


TRANSFORMATION   OF  CO-ORDINATES. 


51 


If  we  refer  it  to  Y"  and  X",  the  equation  assumes  the  yet 
simpler  form 

y"  =  0. 

Hence,  it  appears  that  the  position  of  the  axes  materially 
affects  the  form  of  the  equation  of  a  locus  referred  to  them. 

Note.  —  The  equation  of  a  locus  which  is  referred  to  rec- 
tangular co-ordinates  is  called  the  Eectangular  Equation  of 
the  locus ;  when  referred  to  polar  co-ordinates,  the  equation  is 
called  the  Polar  Equation  of  the  locus. 

32.  To  find  the  equation  of  transformation  from  one  system 
of  co-ordinates  to  a  ])arallel  system,  the  origin  being  changed. 


Y 

Y' 

"v 

V 

\. 

^M 

o' 

A 

0 

[ 

)         \ 

I 

Fig.  18. 

Let  CM  be  any  plane  locus  referred  to  X  and  Y  as  axes, 
and  let  P  be  any  point  on  that  locus.  Draw  PB  ||  to  OY ; 
then  from  the  figure,  we  have, 

(OB,  BP)  =  (x,  y)  for  the  co-ordinates  of  P  when  referred 
to  X  and  Y  ; 

(O'A,  AP)  =  (x',  y')  for  the  co-ordinates  of  P  when  referred 
to  X'  and  Y' ; 

(OD,  DO')  =  (a,  b)  for  the  co-ordinates  of  0',  the  new 
origin. 


52  PLANE  ANALYTIC   GEOMETRY. 

From   the  figure    OB  =  OD  +  DB  ;  and  BP  =  BA  +  AP  ; 

hence  x  =  a  -\-  x'  and  y  —  h  -\-  y' 

are  the  desired  equations. 

As  these  equations  express  the  relations  between  x,  a,  x',  y, 
b,  and  y'  for  any  point  on  the  locus  they  express  the  relations 
between  the  quantities  for  every  point.  Hence,  since  the 
equation  of  th'^  locus  CM  expresses  the  relationship  between 
the  co-ordinates  of  every  point  on  it  if  we  substitute  for 
X  and  y  in  that  equation  their  values  in  terms  of  x'  and  y' 
the  resulting  or  transformed  equation  will  express  the  rela- 
tionship between  the  x'  and  y'  co-ordinates  for  every  point 
on  it. 

EXAMPLES. 

1.  What  does  the  equation  y  =  3  ic  +  1  become  when  the 

origin  is  removed  to  (2,  3)  ? 

Ans.   y  ^  o  X  -{-  4:. 

2.  Construct  the  locus  of  the  equation  2  y  —  x  =  2.  Trans- 
fer the  origin  to  (1,  2)  and  re-construct. 

3.  The  equation  of  a  curve  is  y'^  -{-  x"^  -{-  Ay  —  Ax  —  8  =  0; 
what  does  the  equation  become  when  the  origin  is  taken  at 

Ans.    x^  -\-  y  "  =  lb. 

4.  What  does  the  equation  y"^  —  2x'^  —  2y-\-Qx  —  'd  =  ^ 
become  when  the  origin  is  removed  toi  - ,  1  j  ? 

Ans.    2y^  —  4:x'=—  1. 

5.  The  equation  of  a  circle  is  x^  -\-  y^  =  a^  when  referred 
to  rectangular  axes  through  the  centre.  What  does  this 
equation  become  when  the  origin  is  taken  at  the  left-hand 
extremity  of  the  horizontal  diameter  ? 


TRANSFORMATION  OF  CO-ORDINATES. 


53 


33.    To  find  the  equations  of  transforvuition  from  a  rectangu- 
lar system  to  an  obl'ujue  system,  the  origin  being  changed. 


Y 

y 

c 

( 

/     \ 

p 

/     / 

\m^^-""'^ 

/  3^ 

F 

^-' 

^ 

1 

N 

K 

^-^^ 

1 

.^-"ie             0 

h 

D  L     B  ^ 

Fig.  19. 

Let  P  be  any  point  on  the  locus  CM. 

Let  O'Y',  O'X'  be  the  new  axes,  making  the  angles  <jd  and  Q 
with  the  X-axis.     Draw  PA  ||  to  the  Y'-axis ;  also  the  lines 

O'D,  AL,  PB  II  to  the  Y-axis,  and 
AF,  O'K  II  to  the  X-axis. 
From  the  figure,  we  have, 

OB  ^  OD  +  O'N  +  AF,  and 
PB  =  DO'  +  AN  +  PF. 

But  OB  =  X,  OD  =  a,  O'N  =  x!  cos  B,  AF  =  /  cos  cp, 

PB  =  y,  DO'  =  ^>,  AN  =  x'  sin  Q,  PF  =  /  sin  <f ; 

hence,  substituting,  we  have, 

X  =  a  -\-  x'  cos  B  -\- 1/  cos  qp 
y  ^  h  -\-  x'  sin  6  -\-  y'  sin  go 

for  the  required  equations. 

CoR.  1.    If  a  =  0,  and  i  =  0,  0'  coincides  with  0,  and  we 
have, 


(1) 


X  =^  x'  cos  6  -\-  y'  cos  (p  ) 


...  (2) 


y  =  x'  sin  Q  -\-  y'  sin  go  j 
for  the  equations  of  transformation  frovi  a  rectangular  system 
to  an  oblique  system,  the  origin  remaining  the  same. 


54  PLANE  ANALYTIC  GEOMETRY. 

Cor.  2.  lta  =  0,b  ==0,  and  (p  =  90°  +  6,  0'  coincides  with 
0  and  the  new  axes  X'  and  Y'  are  rectangular.  Making  these 
substitutions,  and  recollecting  that 

cos  (p  =  cos  (90°  -\-  0)  =  —  sin  6,  and 
sin  (p  =  sin  (90°  -{-  0)  =  cos  0, 


we  have, 


X  =  x'  cos  6  —  y'  sin  6  \  .ov 

^  =  ic'  sin  6  -\-  y'  cos  ^  i 


yb?"  i/ie  equations  of  transformation  from  one  rectangular  system 
to  another  rectangular  system,  the  origin  remaining  the  same. 

Note.  —  If  we  find  the  values  of  x'  and  y'  in  equations  (2) 
in  terms  of  x  and  y  we  obtain  the  equations  of  transforma- 
tion from  an  oblique  system  to  a  rectangular  system,  the 
origin  remaining  the  same. 


EXAMPLES. 

1.  What  does  the  equation  x^  -f  y'^  =  16  become  when  the 
axes  are  turned  through  an  angle  of  45°  ? 

Ans.    The  equation  is  unchanged. 

2.  The  equation  of  a  line  {^  y  =  x  —  1 ;  required  the  equa- 
tion of  the  same  line  when  referred  to  axes  making  angles  of 
45°  and  135°  with  the  old  axis  of  x. 

Ans.     y  =  —  V?- 

3.  What  does  the  equation  of  the  line  in  Example  2  become 
when  referred  to  the  old  Y-axis  and  a  new  X-axis,  making  an 
angle  of  30°  with  the  old  X-axis. 

Ans.     2y  =  (V3  -  1)  a;  -  2. 


TRANSFORMATION  OF  CO-ORDINATES. 


55 


34.    To  find  the  equations  of  transformation  from  a  rectangu- 
lar system  to  a  polar  system,  the  origin  and  pole  non-coincident. 


Y 

C' — 

7 

P 

S 

•  ^ 

l 

F 

^M 

i 

1 

.^'^ 

1 

^y 

0 

1 

1 1 — 

Fig.  20. 

Let  O'  {a,  b)  be  the  pole  and  O'S  the  initial  line,  making  an 
angle  cp  with  the  X-axis.  Let  CM  be  any  locus  and  P  any 
point  on  it.     From  the  figure,  we  have, 

OB  =  OD  +  O'F, 
BP  =  DO'  +  PP. 

But  OB  =  a;,  OD  =  a,  O'F  =  O'P  cos  PO'F  =  r  cos  {$  +  (?) 
BP  =  y,  DO'  =  Z>,  FP  =  O'P  sin  PO'F  =  r  sin  (^  +  cp)  ; 

hence,  substituting,  we  have, 


X  =  a  -\-  r  cos  (6  -\-  (f))\ 
y  =  b  -\-  r  sin  (^  +  <p  )  ) 


(1) 


for  the  required  equations. 

CoR.  1.    If  the  initial  line  O'S  is  parallel  to  the  X-axis  (it  is 
usual  to  so  take  it)  qp  =  0,  and 


X  =  a  -\-  r  cos 
y  =  b  -\-  r  sin  ( 


(2) 


become  the  equations  of  transformation. 

CoR.  2.    If  the  pole  is  taken  at  the  origin  0,  and  the  initial 
line  made  coincident  with  the  X-axis  a.  =  0,  6  =  0,  and  go  =  0. 


56  PLANE  ANALYTIC  GEOMETRY. 

Hence,  in  this  case, 

X  =  r  COB  6  >  /o\ 

y  ^  r  siu  0  ) 

will  be  the  required  equations  of  transformation. 

35.  To  find  the  equations  of  transformation  from  a  polar 
system  to  a  rectangular  system. 

1°.  When  the  pole  and  origin  are  coincident,  and  when  the 
initial  line  coincides  with  the  X-axis. 

From  equations  (3),  Art.  34,  we  have,  by  squaring  and 
adding  r"^  =  x^  -\-  y^ ;  and, 

by  division         tan  0  r=y-  . 

X 

for  the  required  equations.     We  have,  also,  from  the  same 
equations, 

cos  ^  =  -  =         ^         :  sin  ^  =  ^  = y 


2°.  When  the  pole  and  origin  are  non-coinciderit,  and  when 
the  initial  line  is  parallel  to  the  X-axis. 

From  equations  (2)  of  the  same  article,  we  have,  by  a  simi- 
lar process, 

r^  =  (a;  —  a)"  -\-  {y  —  by 
tan.  6  =  ^L^^  j  also 


cos  d  = 


X  —  a 
X  —  a 


r  V(x  -  af  +  {y-  by  ; 


sin  6 


__y  —b y  —  b 


^-        V(a;  —  ay  +  (y  —  by 
for  the  required  equations. 


TRANSFORMATION  OF  CO-ORDINATES.  57 


EXAMPLES. 

1.  The  rectangular  equation  of  the  circle  is  x^-  -{-  7/  =  a^  ■ 
what  is  its  polar  equation  when  the  origin  and  pole  are  coin- 
cident and  the  initial  line  coincides  with  the  X-axis  ? 

Ans.    r  =  a. 

2.  The  equation  of  a  curve  is  (x^  +  2/^)  "  =  *"  (^^  ~  V"^)  '■>  es- 
quired its  polar  equation,  the  pole  and  initial  being  taken  as 

in  the  previous  example. 

Ans.   r^  =  a^  cos  2  0. 

Deduce  the  rectangular  equation  of  the  following  curves, 
assuming  the  origin  at  the  pole  and  the  initial  line  coincident 
with  the  X-axis. 

3.  r  =  a  tan  ^  0  sec  6  5.    r^  ^  a^  sin  2  0 

3.         1 
Ans.     x'^  ^  a- y.  Ans.     (x^ -\- y'^Y  =  2  a^xy. 

4.  r^  =  a^  tan  6  sec^  6  6.    r  =  a  (cos  0  —  sin  0) 
Ans.    x^  =  a^y.  Ans.    x^  -{-  y^  ^  a  (x  —  y) 

GENERAL  EXAMPLES. 

Construct  each  of  the  following  straight  lines,  transfer 
the  origin  to  the  point  indicated,  the  new  axes  being  parallel 
to  the  old,  and  reconstruct  : 

1.  ?/  =  3  cc  -f  1  to  (1,  2).  5.  y  =  sx  -{-  b  to  (c,  d). 

2.  2  y  -  a;  -  2  =  0  to  (-  1,  2).     6.  y  -\-  2  x  =  0  to  (2,  —  2). 

3.  i  ?/  +  X  —  4  =  0  to  (-  2,  —  1).  7.  y  =  nix  to  (Z,  n). 

4.  y -f  ^ -}- 1  =  0  to  (0,  2).  8.  ?/ —  4x -f  c  =  Oto(cZ,o). 

What  do  the  equations  of  the  following  curves  become  when 
referred  to  a  parallel  (rectangular)  system  of  co-ordinates 
passing  through  the  indicated  points  ? 

9.    3  ^2  +  2  ?/2  =  6,  (V2,  0). 
11.    9  2/^ -4:^2  =-36  (3,0).     ^^-y-^^H       2'^ 


68  PLANE  ANALYTIC   GEOMETRY. 

13.  What  does  the  equation  x^  -{-  y'^  =  4:  become  when  the 
X-axis  is  turned  to  the  left  tlirough  an  angle  of  30°  and  the 
Y-axis  is  turned  to  the  right  through  the  same  angle  ? 

14.  What  does  the  equation  x'^  —  ?/2  _  ^2  become  when  the 
axes  are  turned  through  an  angle  of  —  45°  ? 

15.  What  is  the  polar  equation  of  the  curve  3/^  =  2  px,  the 
pole  and  origin  being  coincident,  and  the  initial  line  coincid- 
ing with  the  X-axis  ? 

16.  The  polar  equation  of  a  curve  is  r  =  a  (1  -|-  2  cos  6)  ; 
required  its  rectangular  equation,  the  origin  and  pole  being 
coincident  and  the  X-axis  coinciding  with  the  initial  line. 

Ans.    (x^  -\-  y^  —  2  ax)^  =  a^  (x^  -f  y-). 

Required  the  rectangular  equation  of  the  following  curves, 
the  pole,  origin,  initial  line,  and  X-axis  being  related  as  in 
Example  16  .. 

17.  r2  = ^  .  20.    r  =  a  sec^  -^ 


Ans.    x^  —  ?/^  =  al 

18.  r  =  a  sin  0.  21.    r  =  a  sin  2  e. 

19.  r  =  ae.  22.    r^  -  2  r  (cos  ^  +  V3  sin  0)  =  5. 

Find   the    polar   equations  of   the  loci  whose  rectangular 
equations  are : 

23.  x^  =  y^(2a  —  x).       25.    aY  =  «-^*  -  ^^• 

24.  4.a^x  =  y-  (2  a-  x).  26.    x^  ^  y^  =  a^. 


THE   CIRCLE. 


59 


CHAPTER  V. 


THE    CIRCLE. 


36.  The  circle  is  a  curve  generated  by  a  point  moving  in 
tlie  same  plane  so  as  to  remain  at  the  same  distance  from  a 
fixed  point.  It  will  be  observed  that  the  circle  as  here  de- 
fined is  the  same  as  the  circumference  as  defined  in  plane 
geometry. 

37.  Given  the  centre  of  a  circle  and  its  radius  to  deduce  its 
equation. 

Y 


Fig.  21. 


Let  C  (x',  y')  be  the  centre  of  the  circle,  and  let  P  be  any 
point  on  the  curve.  Draw  CA  and  PM  ||  to  OY  and  CIST  ||  to 
OX;  then 

(OA,  AC)  =  (x',  y')  are  the  co-ordinates  of  the  centre  C. 
(OM,  MP)  =  (x,  y)  are  the  co-ordinates  of  the  point  P. 


60  PLANE  ANALYTIC   GEOMETRY. 

Let  CF  =  a.     From  the  figure,  we  have, 
CN2  +  NP2  =.  CP2 ;  ...  (1) 

But  QW  =  (OM  -  0A)2  ={x-  x'Y, 

NP2  =  (MP  -  ACy  =  (y-  y'f,  and 
CP2  =  a" 

Substituting  tliese  values  in  (1),  we  have, 

(^  _  :^'y  +  (y  _  y'y  =  «2  .  .  .   (2) 

for  the  required  equation.  For  equation  (2)  expresses  the 
relation  existing  between  the  co-ordinates  of  any  point  (P)  on 
the  circle ;  hence  it  expresses  the  relation  between  the  co- 
ordinates of  every  point.  It  is,  therefore,  the  equation  of  the 
circle. 

If  in  (2)  we  make  a?'  =  0  and  y'  =  0,  we  have, 
x^  -^y''  =  a'  .  .  .  (3) 

or,  SAanmetrically, \-  ^  =  1  .  .  .   (4) 

a^        a^ 

for  the  equation  of  the  circle  when  referred  to  rectangular 
axes  passing  through  the  centre. 

Let  the   student  discuss  and  construct  equation  (3).     See 
Art.  11. 

Cor.  1.    If  we  transpose  cc^  in  (3)  to  the  second  member  and 
factor,  we  have, 

y"^  =  (a  -\-  x)    (a  —  x)  ; 

i.e.,  in  the  circle  the  ordinate  is  a  mean  "projiortional  between 
the  segments  into  lohich  it  divides  the  diameter. 

CoR.  2.    If  we  take  £,  Fig,  21,  as  the  origin  of  co-ordinates, 
and  the  diameter  Z,H  as  the  X-axis,  we  have, 

£0  =  x'  =  a  and  ?/'  =  0. 

These  values  of  x'  and  y'  in  (2)  give 

{x  —  ay  -|-  ?/2  =  a^, 

or,  after  reduction,  x^  -{-  y"^  —  2  ace  =  0  .  .  .   (5) 

for  the  equation  of  the  circle  when  referred  to  rectangular 


THE   CIRCLE.  61 

axes    taken   at   the   left   hand   extremity    of   the    horizontal 
diameter. 

38.  Every  equation  of  the  second  degree  between  two  varia- 
bles, in  which  the  coefficients  of  the  second  jiowers  of  the 
variables  are  equal  and  the  term  in  xy  is  missing,  is  the  equa- 
tion of  a  circle. 

The  most  general  equation  of  the  second  degree  in  which 
these  conditions  obtain  is 

ax^  -\-  aif  -\-  ex  -\-  dy  -\-  f  =  ^ .  .  .  .   (1) 
Dividing  through  by  a  and  re-arranging,  we  have, 

x^  -\-  -  X  -{-  y-  ^  -  y  =  —-'-  . 
a  a  a 

If  to  both  members  we  now  add 
c2  d^ 


4  a^        4  (^2  ' 

the  equation  may  be  put  under  the  form 

,      cW    f      .     d  Y      c^  +  d^-4.af 
2  a  J        \  2  a  J  4za^ 

Comparing  this  with  (2)  of  the  preceding  article,  we  see 
that  it  is  the  equation  of  a  circle  in  which 

G  d 

2^'  "~2^ 

are  the  co-ordinates  of  the  centre  and 


^  c^  -\- d^  -  4.  af  ig  the  radius. 
2a 

Cor.  1.  If  ax  ^  +  ay^  -\-  cx  -\-  dy  -\-  m  =  0  be  the  equation 
of  another  circle,  it  must  be  concentric  with  the  circle  repre- 
sented by  (1)  ;  for  the  co-ordinates  of  the  centre  are  the  same. 
Hence,  when  the  equations  of  circles  have  the  variables  in 


62  PLANE  ANALYTIC  GEOMETRY. 

their  terms  affected  with  equal  coefficients,  each  to  each,  the 
circles  are  concentric.     Thus 

2  X-  -^  2  if  +  3  X  +  4.tj  -{-  25  =  0 
are  the  equations  of  concentric  circles. 


EXAMPLES. 

What   is   the    equation    of   the    circle   when  the  origin  is 
taken. 

1.  At  D,  Fig.  21  ?  Ans.     x"  ^  y"  —  2  ay  =  0. 

2.  At  K,  Fig.  21  ?  Ans.     x"^  ■^y'^  +  2  ay  =  0. 

3.  At  H,  Fig.  21  ?  Ans.     x-  +  y-  +  2  ax  =  {). 

What  are  the  co-ordinates  of  the  centres,  and  the  values  of 
the  radii  of  the  following  circles  ? 

4.  4a;2  +  4  2/2-8x-8?/  +  2  =  0. 

Ans.     (1,  1),  a  =  yi 

5.  ^2  ^  2/^  -I-  4  cc  _  6  ?/  —  3  =  0. 

Ans.     (-  2,  3),  a  =  4. 

6.  2  x2  +  2  2/2  _  8  a;  =  0. 

Ans.     (2,  0),  a  =  2. 

7.  ^2  _^  2/2  _  6  ^  ^  0. 

Ans.     (3,  0),  a  =  3. 

8.  a;2  +  ?/2  —  4  X  +  8  ?/  —  5  =  0. 

^?zs.     (2,  —  4)  a  =  5. 

9.  ic^  -|-  2/^  —  '''^^  -\-  n  y  -\-  c  =  0. 

10.  a;'^  +  2/^  =  ■'^• 

11.  x^  —  ^x  =  —  y-  —  my. 

12.  ^2  +  y2  ^  c2  +  <Z2_ 

13.  x"^  -\-  ex  -{-  y"^  =  f. 


THE   CIRCLE. 


63 


Write  the  equations  of  the  circles  whose  radii  and  whose 
centres  are 

14.  «  =  3,  (0,  1).  18.    a  =  m,  (b,  c). 
Ans.   x^  -\-  y'^  —  2  7/  ^  8. 

15.  a  =  2,  (1,  -  2).  19.    a  =  b,  (e,  -  d). 
Ans.     X-  -{-  y"^  —  2  X  -\-  4:  y  -\- 1  =0. 

16.  a  =  5,  (-  2,  -  2).  20.    a  =  5,  {I,  k). 

Afis.     X-  -\-  y'^  -\-  4:  y  -\-  4z  X  =  17 . 

17.  a  =  4,  (0,  0).  21.    a  =  k,  (2,  h). 
Ans.     x~  -\-  y'  =  16. 

22.  The  radius  of  a  circle  is  5 :  what  is  its  equation  if  it  is 
concentric  with  cc^  +  ^Z^  —  4ic=2? 

Ans.     x^  -\-  y^  —  4:  X  =  21. 

23.  Write  the   equations  of   two  concentric  circles  which 
have  for  their  common  centre  the  point  (2,  —  1). 

24.  Find  the  equation  of  a  circle  passing  through   three 
given  points. 

39.    To  deduce  the  polar  equation  of  the  circle. 


The  equation  of  the  circle  when  referred  to  OY,  OX  is 
(X  -  x'Y  +  (y  ~  y'Y  =  a\ 


64  PLANE  ANALYTIC  GEOMETRY. 

To   deduce  the  polar  equation  let  P  be  any  point  of  the 
curve,  then 

(OA,  AP)  =  {X,  y) 
(OB,  BO')  =  {x',  y') 
(OP,  POA)  =  {r,  6) 
(00',  O'OB)  =  (/,  6') 

From  the  figure,  OA  =  x  =  r  cos  6,  AP  ^  y  =  r  sin  9, 

OB  =  cc'  =  r'  cos  0',  BO'  =  ^/  =  r'  sin  &  ; 

hence,  substituting,  we  have, 

(?•  cos  6  —  r'  cos  6'y  +  (^ sin  ^  —  ?''  sin  6'Y  =  a^. 
Squaring  and  collecting,  we  have, 

r2(cos2  0  +  sin^  0)  +  r'  ^(cos-  6'  +  sin-  ^')  —  2  rr' (cos  ^  cos  0' 
-\-  sin  ^  sin  6')  =  o? 
i.e.,  r^  -f  r'2  -  2  rr'  cos  (^  -  e')  =  a^  _   .  _   ^-l>) 

•IS  the  polar  equation  of  the  circle. 

This  equation  might  have  been  obtained  directly  from  the 
triangle  OO'P. 

CoR.  1.    If  6'  =  0,  the  initial  line  OX  passes  through  the  cen- 
tre and  the  equation  becomes 

^2  _j_  j/2  _  2  rr'  cos  0  =  a^. 

Cor.  2.    If  &  =  0,  and  /  =  a,  the  pole  lies  on  the  circum- 
ference and  the  equation  becomes 

r  =  2  a  cos  9. 

CoE.  3.    If  9'  =  0,  and  r'  =  0,  the  pole  is  at  the  centre  and 
the  equation  becomes 


40.  To  show  that  the  supplemental  chords  of  the  circle  are 
perpendicular  to  each  other. 

The  supplemental  chords  of  a  circle  are  those  chords  which 
pass  thi'ough  the  extremities  of  any  diameter  and  intersect  each 
other  on  the  circicmference. 


65 


BIG.  23. 


Let  PB,  PA  be  a  pair  of  supplemental  chords.     We  wish  to 
prove  that  they  are  at  right  angles  to  each  other. 
The  equation  of  a  line  through  B  (—  «,  o)  is 

y  =  s  (x  +  a). 
For  a  line  through  A  (a,  o),  we  have 

y  ^  s'  (x  —  a). 
Multiplying  these,  member  by  member,  Ave  have 
y^  =  ss'  (x-  —  rt.^)  .  .  .   (ci) 
for  an   equation  which  expresses   the    relation    between  the 
co-ordinates  of  the  point  of  intersection  of  the  lines. 

Since  the  lines  must  not  only  intersect,  but  intersect  07i  the 
circle  whose  equation  is 

y^  =  ^2  —  ^2^ 

this  equation  must  subsist  at  the  same  time  with  equation  (a) 
above  ;  hence,  dividing,  we  have 

1  =  —  ss', 
or,  1  +  ss'  =  0  .  .  .  (1) 

Hence  the  supplemental  chords  of  a  circle  are  perpendicular 
to  each  other. 

Let  the  student  discuss  the  proposition  for  a  pair  of  chords 
passing  through  the  extremities  of  the  vertical  diameter. 


66 


PLANE  ANALYTIC  GEOMETRY. 


41.    To  deduce  the  equation  of  the  tangent  to  the  circle. 


Fig.  24. 


Let  CS  be  any  line  cutting  the  circle  in  the  points  P'  (x',  y')^ 
P"  {x",  y").     Its  equation  is 

y-y'  -  'i'  I  'C'  (^  -  ^')-  (^^*-  26,  (4) ). 

Since  the  points  {x',  y'),  (x",  y")  are  on  the  circle,  we  have 
the  equations  of  condition 

x""  +  ?/'"  =  ft"  ...   (1) 
x"^  +  /'2  =  a^  .  .  .   (2) 

These  three  equations  must  subsist  at  the  same  time ;  hence, 
subtracting  (2)  from  (1)  and  factoring,  we  have, 

(^'  +  ^")  (^'  -  ^")  +  {y'  +  y")  {y'  -  y")  =  o ; 

y'  —  y"  ^        x'  +  x" 
"  x'  —  x"  ~        /  +  y"  ' 
Substituting  in  the  equation  of  the  secant  line  it  becomes 

x'  +  x" 


y  -y 


(x  —  x') 


(3) 


y'  +  y' 

If  we  now  revolve  the  secant  line  upward  about  P"  the 
point  P'  will  approach  P"  and  will  finally  coincide  with  it 
when  the  secant  CS  becomes  tangent  to  the  curve.     But  when 


THE   CIRCLE.     .  67 

F'  coincides  with  V",  x'  =  x"  and  ?/  =  y";  hence,  substituting 
in  (3)  we  have, 

y-.f^-'^ix-x"),    ...    (4) 

or,  after  reduction, 

xx"  +  yij"  =  a?;  .  ..   (5) 

or,  symmetrically, 

^^     I  yy   =  1  ...  (6) 

or  a^ 

for  the  equation  of  the  tangent. 

ScHOL.  The  Sub-tangent  for  a  given  point  of  a  curve  is  the 
distance  from  the  foot  of  the  ordinate  of  the  point  of  tangency 
to  the  point  in  which  the  tangent  intersects  the  X-axis ;  thus, 
in  Fig.  24,  AT  is  the  sub-tangent  for  the  point  F".  To  find 
its  value  make  y  =  0  in  the  equation  of  the  tangent  (5)  and 
we  have, 

OT  ==  cc  =  —  . 

x" 

But  AT  =  OT  -  OA  ^  ^  -  x" 

x" 

.-.    sub-tangent  = ; =  ^  . 

x''  x ' 

42.    To  deduce  the  equation  of  the  normal  to  the  circle. 

The  normal  to  a  curve  at  a  given  point  is  a  line  perpen- 
dicular to  the  tangent  drawn  at  that  point. 

The  equation  of  any  line  through  the  point  F"  (x",  y")  Fig. 
24,  is  y  -  2j"  =  s  (x  -  x")  ...   (1) 

In  order  that  this  line  shall  be  perpendicular  to  the  tangent 
F"T,  we  must  have 

1  -j_  S5'  =  0. 

x"  v" 

But  Art.  41,  (4)  s'  = — ;  hence,  we  must  have  s  =  -i—  . 

y"  x" 


68  PLANE  ANALYTIC   GEOMETRY. 

Therefore,  substituting  in  (1),  we  have, 

y  -y"  =  ^'^ri^  -  ^")  •  •  •  (2) ; 

X 

or,  after  reduction, 

yx"  -  xy"  =  0  ...  (3) 

for  the  equation  of  the  normal. 

We  see  from  the  form  of  this  equation  that  the  normal  to 
tlie  circle  passes  through  the  centre. 

ScHOL.  The  SuB-NOKMAL  for  a  given  point  on  a  curve  is 
the  distance  from  the  foot  of  the  ordinate  of  the  point  to  the 
point  in  which  the  normal  intersects  the  X-axis.  In  the  circle, 
we  see  from  Fig.  24  that  the 

Sub-normal  =  x" . 

43.  By  methods  precisely  analogous  to  those  developed  in 
the  last  two  articles,  we  may  prove  the  equation  of  the  tangent 
to 

ix  —  x'y  +  (^  —  y'Y  =  a} 
to  be 

{X  -  x')  ix"  -  x')  +  {y-  y')  {y"  -  y')  =  a?  .  .  .   (1) 
and  that  of  the  normal  to  be 

{y  -y")  {x"  -  x')  -  (^  -  ^")  {y"  -y')  =  o  .  .  .  (2) 

Let  the  student  deduce  these  equations. 

EXAMPLES. 

1.  What  is  the  polar  equation  of  the  circle  ax^  +  ay^  -\-  ex  -\- 
dy  +/=  0,  the  origin  being  taken  as  the  pole  and  the  X-axis 
as  the  initial  line  ? 

Ans.     ?'^  +  (  -  cos  ^  -| —  sin  ^  )?•-)-  =^  =  0. 


^a  a  a 

2.  What  is  the  equation  of  the  tangent  to  the  circle 
x^  -\-  y^  =  25  at  the  point  (3,  4)  ?  The  value  of  the  sub- 
tangent  ?  A71S.     3x  -\-  4,y  =  25;  J/. 

3.  What  is  the  equation  of  the  normal  to  the  circle 
x^  -\-  y^  =  37  at  the  point  (1,  6)  ?  What  is  the  value  of  the 
sub-normal  ?  Ans.    y  =  6x;  1. 


THE   CIRCLE.  69 

4.  What  are  the  equations  of  the  tangent  and  normal  to  the 
circle  a?^  +  t/^  =  20  at  the  point  whose  abscissa  is  2  and  ordi- 
nate negative  ?  Give  also  the  values  of  the  sub-tangent  and 
sub-normal  for  this  point. 

Ans.    2x  -4:ij  =  20;     2?/  +  4x  =  0; 
Sub-tangent  =  8  ;    sub-normal  =  2. 

Give  the  equations  of  the  tangents  and  normals,  and  the 
values  of  the  sub-tangents  and  sub-normals,  to  the  following 
circles : 

5.  x^-{-i/  =  12,  at  (2,  +  V8). 

6.  a;2  4-  2/2  =  25,  at  (3,  -  4). 

7.  X-  -{-if  =  20,  at  (2,  ordinate  +). 

8.  x^  -\-  f  =  32,  at  (abscissa  -f,  —  4). 

9.  x^  -{-  f  =  a^,  Sit  (b,  c). 

10.  x'^  -\-  f  =  m,  at  (1,  ordinate  -|-). 

11.  ic-  -)-?/-  =  k,  at  (2,  ordinate  — ). 

12.  :k-  -f-  y2  =  18,  at  (m,  ordinate  -(-), 

13.  Given  the  circle  x^  -\-  y^  =  45  and  the  line  2-1/  -\-  x  =  2; 
required  the  equations  of  the  tangents  to  the  circle  which  are 
parallel  to  the  line. 

J  (3x  +  62/  =  45. 

^*-     \3x-\-6y  =  -45. 

14.  What  are  the  equations  of  the  tangents  to  the  circle 
x^  -{-  f  ^=  45  which  are  perpendicular  to  the  line  2  y  -\-  x  =  2? 

,  \oy  —  Qx  =  45. 

\  Q  X  —  S  y  =  45. 

16.  The  point  (3,  6)  lies  outside  of  the  circle  x^  -\-  y^  ^  9; 
required  the  equations  of  the  tangents  to  the  circle  which 
pass  through  this  point. 

,  (  X  =  3. 

^^^-     Uy-3x  =  W. 


70  PLANE  ANALYTIC   GEOMETRY. 

17.  What  is  the  equation  of  the  tangent  to  the  circle 
(x  -  2)2  +  (y  -  3)2  =  5  at  the  point  (4,  4)  ? 

Ans.    2  X  -\-  y  =  12. 

18.  Tlie  equation  of  one  of  two  supplementary  chords  of 

the    circle  a;^  +  y^  =  9  is  t/  =  §  a;  +  2,  what  is  the  equation 

of  the  other  ? 

Ans.    2  ?/  +  3  .X  =  9. 

19.  Find  the  equations  of  the  lines  which  touch  the  circle 
(x  —  a)'^  ^  (fj  —  by  ^=  r-  and  which  are  parallel  to  y  =  sa:  +  c. 

20.  The  equation  of  a  circle  is  a:;^  +  2/^  —  4  x  +  4  ?/  =  9  ; 
required  the  equation  of  the  normal  at  the  point  whose 
abscissa  =  3,  and  whose  ordinate  is  positive. 

Ans.   4:X  —  y  =  10. 

44.  To  find  the  length  of  that  portion  of  the  tangent  lying 
bettveen  any  point  on  it  and  the  point  of  tangency. 

Let  (xi,  yi)  be  the  point  on  the  tangent.  The  distance  of 
this  point  from  the  centre  of  the  circle  whose  equation  is 

(x  —  x'Y  +  (y  —  y'Y  =  o?-  is  evidently 
V(a;,  -  xy  +  (2/1  -  y'f.     See  Art.  27,  (1). 

But  this  distance  is  the  hypothenuse  of  a  right  angled  tri- 
angle whose  sides  are  the  radius  a  and  the  required  distance 
d  along  the  tangent ;  hence 

d'  =  ix,  -  xy-  +  (yi  -  yj  -  a^  ...  (1) 

Cor.  1.    If  x'  =  0  and  y'  =  0,  then  (1)  becomes 
d'  =  x^  +  y,2  _  a^  .  .  .   (2) 
as  it  ought. 

45.  To  deduce  the  equation  of  the  radical  a.xis  of  tivo  given 
circles. 

The  Radical  axis  of  two  circles  is  the  locus  of  a  point 
from  which  tangents  drawn  to  the  two  circles  are  equal. 


THE   CIRCLE.  71 


Fig.  25. 


Let  {x  —  x'Y  +  (i/  —  y'Y  =  a% 

(x  —  x"y  -\-  (jj  —  y")  =  b^  be  the  given  circles. 

Let  P  (xi,  iji)  be  any  point  on  the  radical  axis ;  then  from 
the  preceding  article,  we  have, 


d^  =  (xi  -X'Y  +  (7/1  -  2/') 


l\2 


a^ 


d'-  =  (Xi  —  x"y  +  (yi  -  y"y  —  P 
.-.   by    definition  (x^  —  x')-  +  (yi  —  ?/')-  —  a"  =  (x^  —  x"Y 

+  (2/1  ~  y'Y'  ~  ^^j  hence,  reducing,  we  have, 

2  (a;"  —  x')  a;i  +  2  (?/"  —  y')  y^  =  x'"'^  —  x'-  +  y"^  —  y'^ 
-\-  a-  —  b~. 

Calling,  for  brevity,  the  second  inember  m,  we  see  that 
(iCi,  ?/i)  will  satisfy  the  equation. 

2  {x"  —  x')x  +2  (y"  —  y')y  =  m  .  .  .   (1) 

But  (ccj,  ?/i)  is  awy  point  on  the  radical  axis ;  hence  every 
point  on  that  axis  will  satisfy  (1).  It  is,  therefore,  the  re- 
quired equation. 

CoK.  1.  If  c  =  0  and  c'  =  0  be  the  equation  of  two  circles,, 
then,  c  —  c'  =  0 

is  the  equation  of  their  radical  axis. 


72  PLANE  ANALYTIC  GEOMETRY. 

Cor.  2.  From  the  method  of  deducing  (1)  it  is  easily  seen 
that  if  the  two  circles  intersect,  the  co-ordinates  of  their  points 
of  intersection  must  satisfy  (1) ;  hence  the  radical  axis  of  two 
intersecting  circles  is  the  line  joining  their  points  of  intersection, 
PA,  Fig.  25. 

Let  the  student  prove  that  the  radical  axis  of  any  two 
circles  is  perpendicular  to  the  line  joining  their  centres. 

46.  To  shoio  that  the  radical  axes  of  three  given  circles  in- 
tersect in  a  common  ■point. 

Let  c  =  0,  c'  =  0,  and  c"  =  0 

be  the  equations  of  the  three  circles. 

Taking  the  circles  two  and  two  we  have  for  the  equations  of 
their  radical  axes 

c  _  c'  =  0  .  .  .   (1) 

c-c"  =  0  .  .  .  (2) 

c'  -  c"  =  0  .  .  .  (3) 

It  is  evident  that  the  values  of  x  and  i/  which  simultaneously 
satisfy  (1)  and  (2)  will  also  satisfy  (3)  ;  hence  the  proposition. 

The  intersection  of  the  radical  axes  of  three  given  circles  is 
called  The  Kadical  Cejs^tre  of  the  circles. 

EXAMPLES. 

Find  the  lengths  of  the  tangents  drawn  to  the  following 
circles : 


1.  (x  -  2)2  +  {y-  3)-  =  16  from  (7,  2). 

2.  a;'  +  (y  +  2)^  =  10  from  (3,  0). 

3.  (x  —  af  ^y-  =  V1  from  (h,  c). 

4.  a;'  +  ?/-  -  2  ic  +  4  2/  =  2  from  (3,  1). 

5.  ^2  +  2/'  =  25  from  (6,  3). 


Ans.     d  =  VlO. 
Ans.     d  =  V3 . 

Ans.     d  =  V20. 


THE   CIRCLE.  73 

6.  a;2  +  3/2  _  2  x  =  10  from  (5,  2). 

Ans.   d  =  3. 

7.  (x.  -  ay  -^(ij  -hy  =  c  from  {d,  /). 

8.  x2  +  2/2  _  4  7/  =  10  from  (0,  0). 

Give  the  equations  of  the  radical  axis  of  each  of  the  follow- 
ing pairs  of  circles  : 

9.     Ux-  2)2  +  (y  -  3)2  -  10  =  0. 
t(a:+3)2  +  (2/  +  2)^-6  =  0. 

Ans.   5  X  -\-  o  y  -{■  2  =  0. 

10.  J-  a;2  +  2/2  —  4  y  =  0. 

\(x-3y +  t/-^  =  Q.  Ans.    3x  =  2ij. 

11.  r(a;+3)2  +  2/2_2y_8=0. 

\x-  -\-  y-  —  2  y  =  0.  Aris.    x  =  —  J. 

12.  f  (cc  +  ay  +  2/2  _  c2  =  0. 
|:r2  +  (2/-3)2-16  =  0. 

13.  (  a;2  +  2/2  =  16. 

I  (.r  —  1)'  +  r  =  «'• 

14.  j  a;2  +  (2/  -  ay  =  g\ 
X(x-2y  +  y'-  =  d\ 

Find  the  co-ordinates  of  the  radical  centres  of  each  of  the 
following  systems  of  circles  : 

15.  {{x-  3)2  +  2/2  =  16. 

x'  -\-  y'  =  9. 

x2  4- (2/ -  2)2  =  25.  Ans.    (1,-3). 

16.  r  cc2  _j_  2/2  —  4  cc  -)-  6  2/  —  3  =  0, 


■.x--\-y'^  —  4iX^=  12. 

(a;2  +  2/2  +  62/  =  7.  ^ws.    (1, -i). 


17. 


18. 


«^  +  r  =  «^- 

(^  -  1)2  +  f=9. 

x"  -\-y-  —  2x  -\-^y  = 

=  10. 

a;2  -|-  2/^  ""  ^-"^  =  <^- 
a-2  -j-  2/2  =  m. 

^^  +  //"  "~  <^'Z/  =  ^• 

74 


PLANE  ANALYTIC  GEOMETRY. 


47.  To  find  the  condition  that  a  straight  line  y  =  sx  -{- h 
must  fulfil  in  order  that  it  may  touch  the  circle  x'  -\-  y-  =  a\ 

In  order  that  the  line  may  touch  the  circle  the  perpendicu- 
lar let  fall  from  the  centre  on  the  line  must  be  equal  to  the 
radius  of  the  circle. 

From  Art.  21,  Fig.  13,  we  have 

sec  )'       VI  +  tan.-'  y 


P 


Vi  + «' 

hence,  a^  (1  4- s')  =  b''  .  .  .   (1) 

is  the  required  condition. 

CoR.  1.    If  we  substitute  the  value  of  b  drawn  from  (1)  in 
the  equation  y  =  sx  -\-  b,  we  have 

y  ^  sx  -^a  Vl  +  s'^  .  .  .  (2) 
for  the  equation  of  the  tangent  in  terms  of  its  slope. 

48.    Two  tangents  are  drawn  from  a-  point  ivithout  the  circle  ; 
requii'ed  the  equation  of  the  chord  joining  the  points  of  tangency. 


Fig.  26. 


Let  P'  {x',  y')  be  the  given  point,  and  let  P'P",  PT,  be  the 
tangents  through  it  to  the  circle. 


THE   CIRCLE.  75 

It  is  required  to  deduce  the  equation  of  PP''. 
The  equation  of  a  tangent  through  P"  (x",  y")  is 
xx"    .    yy"  _  -I 

Since  P'  (x',  y')  is  on  this  line,  its  co-ordinates  must  satisfy 
tlie  equation ;  hence 

^'^''  I  y'v"  ^  1 

The  point  {x",  y"),  therefore,  satisfies  the  equation 
^  +  ^  =  1;  •■•(!) 

.-.  it  is  a  point  on  the  locus  represented  by  (1).  A  similar 
course  of  reasoning  will  show  that  P  is  also  a  point  of  this 
locus.  But  (1)  is  the  equation  of  a  straight  line  ;  hence,  since 
it  is  satisfied  for  the  co-ordinates  of  both  P"  and  P,  it  is  the 
equation  of  the  straight  line  joining  them.  It  is,  therefore, 
the  required  equation. 

49.  A  chord  of  a  given  circle  is  revolved  about  one  of  its 
points ;  required  the  equation  of  the  locus  generated  by  the 
point  of  intersection  of  a  pair  of  tangents  drawn  to  the  circle  at 
the  points  in  which  the  chord  cuts  the  circle. 

Let  P'  (x',  i/),  Fig.  27,  be  the  point  about  which  the  chord 
P'AB  revolves.  It  is  required  to  find  the  equation  of  the 
locus  generated  by  Pi  {xi,  yi),  the  intersection  of  the  tangents 
APi,  BPi,  as  the  line  P'AB  revolves  about  P'. 

Prom  the  preceding  article  the  equation  of  the  chord  AB  is 


^1^    I   yiV  __  1 


a 


Since  P'  (x',  y')  is  on  this  line,  we  have 


^\^    I  yai 


+ 

a'  a' 

hence  ^  +  "^  =  1  •  •  •  (1) 


a^         a^ 


76 


PLANE  ANALYTIC  GEOMETRY. 


is  satisfied  for  the  co-ordinates  of  Pi  (x^,  y^  ;  hence  Pj  lies  on 
the  locus  represented  by  (1).  But  Pi  is  the  intersection  of 
any  pair  of  tangents  drawn  to  the  circle  at  the  points  in 


Fig.  27. 


which  the  chord,  in  any  position,  cuts  the  circle ;  hence  (1) 
will  be  satisfied  for  the  co-ordinates  of  the  points  of  intersec- 
tion of  every  pair  of  tangents  so  drawn. 

Equation  (1)  is,  therefore,  the  equation  of  the  required 
locus.  We  observe  that  equation  (1)  is  identical  with  (1)  of 
the  preceding  article ;  hence  the  chord  PP''  is  the  locus  whose 
equation  we  sought. 

The  point  P'  (x',  y')   is  called  the  pole  of  the  line  PP" 

(^\yly  =  \\  and  the  line  PP"  /^  +  ^  =  1  Vs  called 
\a^         a^  J  ya^         a^  J 

THE  POLAR  of  the  poiut  P'  {x',  y')  with  regard  to  the  circle 


a^       a^ 


=  1. 


THE   CIRCLE.  77 

As  the  principles  here  developed  are  perfectly  general,  the 
pole  may  be  tvithoiit^  on,  or  within  the  circle. 

Let  the  student  prove  that  the  line  joining  the  pole  and  the 
centre  is  perpendicular  to  the  polar. 

]SroTE.  —  The  terms  ^ole  and  i:)olaT  used  in  this  article  have 
no  connection  with  the  same  terms  used  in  treating  of  polar 
co-ordinates,  Chapter  I. 

50.  If  the  polar  of  the  point  P'  (x',y'),  Fig.  27,  passes  through 
Pi  (^i>  Vi),  then  the  polar  of  F^  (x^,  y^)  ivill  pass  through  F' 
(x',  y'). 

The  equation  of  the  polar  to  P'  (x',  y')  is 

x'x    .   y'y -| 

a^         a- 
In  order  that  Pi  {x^,  t/i)  may  be  on  this  line,  we  must  have, 

^'^1    I  y!]h  ^  1, 

tt"'  a 

But  this  is  also  the  equation  of  condition  that  the  point 
P'  (x',  y')  may  lie  on  the  line  whose  equation  is 

^1^  ^^ii/  =  1 

9         I  9 

a'^         a^ 
But  this  is  the  equation  of  the  polar  of  P^  (xi,  y^)  ;  hence 
the  proposition. 

51.  To  ascertain  the  relationship  hetiveen  the  conjugate  diam- 
eters of  the  circle. 

A  pair  of  diameters  are  said  to  he  conjugate  when  they  are 
so  related  that  when  the  curve  is  referred  to  them  as  axes  its 
equation  will  contain  only  the  second  powers  of  the  variables. 

Let  cc-  -f  ?/2  =  a2  ...   (1) 

be  the  equation  of  the  circle,  referred  to  its  centre  and  axes. 
To  ascertain  what  this  equation  becomes  when  referred  to 
OY',  OX',  axes  making  any  angle  with  each  other,  we  must 
substitute  in  the  rectangular  equation  the  values  of  the  old 


78 


PLANE  ANALYTIC  GEOMETRY. 


co-ordinates  in  terms  of  the  new.     Prom  Art.  33,  Cor.  1,  we 

have 

X  =  x'  cos  6  -{-  y'  cos  go 
y  =  a;'' sin  6  -\-  y'  sin  go 

for    tlie    equations   of    transformation.       Substituting   these 
values  in  (1)  and  reducing,  we  have, 

y'^  +  2x'y'  cos  {(f  —  9)  -\-  x'^  =  a^  .  .  .   (2) 


Now,  in  order  that  OY',  OX'  may  be  conjugate  diameters 
they  must  be  so  related  that  the  term  containing  x'y'  in  (2) 
must  disappear ;  hence  the  equation  of  condition, 

cos  (go  —  ^)  =  0 ; 

.-.  cp  _  ^  =  90°,  or  cp  -  e  =  270°. 

The  conjugate  diameters  of  the  circle  are  therefore  perpen- 
dicular to  each  other.  As  there  are  an  infinite  number  of 
pairs  of  lines  in  the  circle  which  satisfy  the  condition  of  being 
at  right  angles  to  each  other,  it  follows  that  in  the  circle  there 
are  an  infinite  number  of  conjugate  diameters. 


THE   CIRCLE.  79 

EXAMPLES. 

1.  Prove  that  the  line  y  =  V3  x  -\-10   touches   the    circle 

.-r'  -\-  y'^  =  25,  and  find  the  co-ordinates  of  the  point  of  tangency. 

/      5—5 
Ans.     Point  of  tangency     _  -^-y/^,  — 

2.  What  must  be  the  value  of  h  in  order  that  the  line 
y  =  2  X  -{-  b  mB,y  touch  the  circle  x^  +  ?/'  =  16  ?  

Ans.     b  =  ^  V80. 

3.  What  must  be  the  value  of  s  in  order  that  the  line 
y  =  sx  —  4:  may  touch  the  circle  x-  -\-  y"^  =  22  _ 

Ans.     s  =  J-  V7. 

4.  The  slope  of  a  pair  of  parallel  tangents  to  the  circle 
X-  -\-  y-  =  1Q>  is  2 ;  required  their  equations. 

Ans.      W  =  2x  +  Vi0. 

]^y  =  2x  —  V80. 

Two  tangents  are  drawn  from  a  point  to  a  circle ;  required 
the  equation  of  the  chord  joining  the  points  of  tangency  in 
each  of  the  following  cases  : 


Ans.  4  X  +  2  ?/  =  Q. 
Ans.  3  a;  -f  4  2/  =  8. 
Ans.    X  -\-o  y  =^  16. 


5.  From  (4,  2)  to  a;-  +  /  =  9. 

6.  From  (3,  4)  to  x""  -\- if  =  8. 

7.  From  (1,  5)  to  x^  +  f~  =  16. 

8.  From  (a,  b)  to  x^  +  ^f  =  c". 

Ans.    ax  -\-  by  ^=  c^. 

What  are  the  equations  of  the  polars  of  the  following  points  : 

9.  Of  (2,  5)  with  regard  to  the  circle  x^  -{-  y-  =  16? 

Ans.     ^+^  =  1. 
16       16 

10.  Of  (3,  4)  with  regard  to  the  circle  x^  -\-  y-  =  9? 

Ans.    S  X  -(-  4  3/  =  9. 


80  PLANE  ANALYTIC   GEOMETRY. 

11.  Of  (a,  h)  with  regard  to  the  circle  x-  -\-  y^  =  m? 

Ans.    ax  -\-  by  =  m.. 

What  are  the  poles  of  the  following  lines  : 

12.  Of  2  a;  +  3  2/  =  5  with  regard  to  the  circle  x"^  -\- y-  =  2?>  2 

Ans.    (10,  15). 

13.  Of  — [-  y  =  4  with  regard  to  the  circle 


o 


16+16  =  ^'  ^-     (2'*)- 

14.    Oi  y  =  sx  -\-  h  with  regard  to  the  circle 


X 

~2 


a^       a^ 


Ans.     (_^    ^ 
h  '     b 


15.    Pind  the  equation  of  a  straight  line  passing  through 
(0,  0)  and  touching  the  circle  x'^  -\-  y^  —  3  x  -{-  4  y  =  0. 

Ans.     y  =  —  X. 
^       4 

GENERAL  EXAMPLES. 

1.  Find  the  equation  of  that  diameter  of  a  circle  which 
bisects  all  chords  drawn  parallel  to  y  :=  sx  -\-  b. 

Ans.    sy  -\-  X  ^  0. 

2.  Eequired  the  co-ordinates  of  the  points  in  which  the 
line  2y  —  x-{-l  =  0  intersects  the  circle 

^^1/1  =  1 
4  "^  4 

3.  Find  the  co-ordinates  of  the  points  in  which  two  lines 
drawn  through  (3,  4)  touch  the  circle 

^  +  i!  =  1 

9        9 

[The  points  are  common  to  the  chord  of  contact  and  the 
circle.] 


THE   CIRCLE.  81 

4.  The  centre  of  a  circle  which  touches  the  Y-axis  is  at 
(4,  0)  ;  required  its  equation. 

Ans.    {x  —  4)-  -\-  y-  =  16. 

5.  Find  the  equation  of  the  circle  whose  centre  is  at  the 
origin  and  to  which  the  line  y  =  x  -\-  S  is  tangent. 

Ans.    1x'  -^Itf  =  9. 

6.  Given  x~  ^  if  =  16  and  (x  —  5)^  +  ?/^  =  4 ;  required  the 
equation  of  the  circle  which  has  their  common  chord  for  a 
diameter. 

7.  Required  the  equation  of  the  circle  which  has  the  dis- 
tance of  the  point  (3,  4)  from  the  origin  as  its  diameter. 

Ans.     x^  -\-  y"'  —  o  X  —  4  ?/  =  0. 

8.  Find  the  equation  of  tlie  circle  which  touches  the  lines 
represented  by  cc  =  3,  ?/  ^  0,  and  y  ■=  x. 

9.  Find  the  equation  of  the  circle  which  passes  through  the 
points  (1,  2),  (-  2,  3),  (-  1,  -  1). 

10.  Required  the  equation  of  the  circle  which  circumscribes 
the  triangle  whose  sides  are  represented  by  3/  =  0,  3  ?/  =  4  a^, 
and  3^/=  —  4a:;-j-6. 

Ans.     xJ  -\-  y^  —  &  X  — 11  ?/  =  0. 

11.  Required  the  equation  of  the  circle  whose  intercepts 
are  a  and  b,  and  which  passes  through  the  origin. 

Ans.     x^  +  y-  —  ax  —  by  =  0. 

12.  The  points  (1,  5)  and  (4,  6)  lie  on  a  circle  whose  centre 
is  in  the  line  y  =  x  —  4 ;  required  its  equation. 

Ans.     2  x^  +  2  y'-  -  17  X  -  y  =  SO. 

13.  The  point  (3,  2)  is  the  middie  point  of  a  chord  of  the 
circle  x'^  -]-?/-  =  16 ;  required  the  equation  of  the  chord. 

14.  Given  x^  -{-  y^  ^  16  and  the  chord  ?/  —  4  a;  =  8.  Show 
that  a  perpendicular  from  the  centre  of  the  circle  bisects  the 
chord. 

15.  Find  the  locus  of  the  centres  of  all  the  circles  which 
pass  through  (2,  4),  (3,  -  2). 


82  PLANE  ANALYTIC   GEOMETRY. 

16.  Show  that  if  the  polars  of  two  points  meet  in  a  third 
point,  then  that  point  is  the  pole  of  the  line  joining  the  first 
two  points. 

17.  Required  the  equation  of  the  circle  whose  sub-tangent 
=  8,  and  whose  sub-normal  =  2. 

Ans.     x^  -\-  y^  =  20. 

18.  Eequired  the  equation  of  ihe  circle  whose  sub-normal 
=  2,  the  distance  of  the  point  in  which  the  tangent  intersects 
the  X-axis  from  the  origin  beinr  =  8. 

A71S.     X-  -\-  1/^  ^  16. 

19.  Eequired  the  conditions  in  order  that  the  circles 
ax^  -f-  ay^  -\-  ex  -\-  dy  -\-  e  =  Q  and  ax-  -f-  ajf  -\-  kyc.  -\- ly  -{-  m  ^  0 
may  be  concentric. 

Ans.    c  ^^  k,  d  =  I. 

20.  Required  the  polar  co-ordinates  of  the  centre  and  the 
radius  of  the  circle 

■    r"^  —  2r  (cos  ^  +  VS  sin  6)  =  5. 

Ans.    (2,  60°)  ;  r  =  3. 

21.  A  line  of  fixed  length  so  moves  that  its  extremities 
remain  in  the  co-ordinate  axes ;  required  the  equation  of  the 
circle  generated  by  its  middle  point. 

22.  Find  the  locus  of  the  vertex  of  a  triangle  having  given 
the  base  =  2a  and  the  sum  of  the  squares  of  its  sides  =  2  b^. 

Ans.     x"^  -\-  y"^  =  h^  —  a^. 

23.  Find  the  locus  of  the  vertex  of  a  triangle  having  given 
the  base  =2  a  and  the  ratio  of  its  sides 

=  —  .  Ans.    A  circle. 

n 

24.  Find  the  locus  of  the  middle  points  of  chords  drawn 
from  the  extremity  of  any  diameter  of  the  circle 

^  _|-  l!_  =  1. 


THE  PARABOLA. 


83 


CHAPTER   VI. 


THE   PARABOLA. 


52.  The  parabola  is  the  locus  generated  by  a  point  moving 
in  the  same  plane  so  as  to  remain  always  equidistant  from  a 
fixed  point  and  a  fixed  line. 

The  fixed  point  is  called  the  Focus ;  the  fixed  line  is  called 
the  Directrix  ;  the  line  drawn  through  the  focus  perpendic- 
ular to  the  directrix  is  called  the  Axis ;  the  point  on  the  axis 
midway  between  the  focus  and  directrix  is  called  the  Vertex 
of  the  parabola. 

53.  To  find  the  equation  of  the  j^arabola,  given  the  focus  and 
directrix. 


R 

Y 

^^„,,--^'^ 

B 

1 

D^.^-""''^ 

^^ 

D 

0 

// 

\  "^                       ' 

^ 

C 

Fig.  29. 


Let  EC  be  the  directrix  and  let  F  be  the  focus.     Let  OX, 
the  axis  of  the  curve,  and  the  tangent  OY  drawn  at  the  vertex 


84  PLANE  ANALYTIC  GEOMETRY. 

0,  be  the  co-ordinate  axes.     Take  any  point  P  on  the  curve 
and  draw  PA  ||.to  OY,  PB  ||  to  OX  and  join  P  and  F.     Then 
(OA,  AP)  =  {x,  y)  are  the  co-ordinates  of  P. 
From  the  right  angled  triangle  FAP,  we  liave 

tf  =  AP2  =  FP2  -  FA^;  ...   (1) 
But  from  the  mode  of  generating  the  curve,  we  have 
FP2  =  BP-  =  (AO  -f  OD)-  =  (x-\-  0D)2, 
and  from  the  figure,  we  have 

FA-  =  (AO  -  0F)2  =  (x-  OF)l 

Substituting  these  values  in  (1),  we  have 

■if  =  (x  +  ODf  -  (x-  0F)2.  .  .  (2) 

Let  DF  =  2),  then  OD  =  OF  =  -g  ;  hence 


y^  =  ix  +^- 


lA"      U  ^P 


X  — 


or,  after  reduction,         ?/^=  2^j).x  ...   (3) 

As  equation  (3)  is  true  for  cmy  point  of  the  parabola  it  is 
true  for  every  point ;  hence  it  is  the  equation  of  the  curve. 

CoR.  1.    If  (x',  y')  and  (x'^,  y")  are  the  co-ordinates  of  any 
two  points  on  the  parabola,  we  have, 

tf  =  2px'  and  y"^  =  2  px" ; 
hence  y'^ :  y"^ ::  x' :  x"  ; 

i.e.,  the  squares  of  the  ordinates  of  any  two  points  on  the  para- 
bola are  to  each  other  as  their  abscissas. 

ScHOL.    By  interchanging  x  and  y,  or  changing  the  sign  of 
the  second  member,  or  both  in  (3),  we  have 

y"^  =  ~  2px  for  the   equation  of  a  parabola  symmetrical 
with  respect  to  X  and  extending  to  the  left  of  Y; 

cc^  =  2  py  for  the  equation  of  a  parabola  symmetrical  with 
respect  to  Y  and  extending  above  X. 

£c^  =  —  2py  for  the    equation  of  a  parabola  symmetrical 
with  respect  to  Y  and  extending  below  X. 

Let   the    student   discuss    each   of   these    equations.      See 
Art.  13. 


THE  PARABOLA.  85 

54.    To  construct  the  parabola,  given  the  focus  and  directrix. 


Fig.  30. 

First  Method.  —  Let  DR  be  the  directrix  and  let  F  be  the 
focus. 

From  F  let  fall  the  perpendicular  FD  on  the  directrix ;  it 
will  be  the  axis  of  the  curve.  Take  a  triangular  ruler  ADC 
and  make  its  base  and  altitude  coincide  with  the  axis  and 
directrix,  respectively.  Attach  one  end  of  a  string,  whose 
length  is  AD,  to  A ;  the  other  end  to  a  pin  fixed  at  F.  Place 
the  point  of  a  pencil  in  the  loop  formed  by  the  string  and 
stretch  it,  keeping  the  point  of  the  pencil  pressed  against  the 
base  of  the  triangle.  Now,  sliding  the  triangle  up  a  straight 
edge  placed  along  the  directrix,  the  point  of  the  pencil  will 
describe  the  arc  OP  of  the  parabola ;  for  in  every  position  of 
the  pencil  point  the  condition  of  its  being  equally  distant 
from  the  focus  and  directrix  is  satisfied.  It  is  easily  seen,  for 
instance,  that  when  the  triangle  is  in  the  position  A'D'C  that 
FP  =  PD'. 

Second  Method.  —  Take  any  point  C  on  the  axis  and  erect 


86  PLANE  ANALYTIC  GEOMETRY. 

the  periDendicular  P'CP.  Measure  the  distance  DC.  With  F 
as  a  centre  and  DC  (=  FP)  as  a  radius  describe  the  arc  of  a 
circle,  cutting  P'CP  in  P  and  P'.  P  and  P'  will  be  points  of 
the  parabola.  By  taking  other  points  along  the  axis  we  may, 
by  this  method,  locate  as  many  points  of  the  curve  as  may  be 
desired. 

55.  To  find  the  Latiis-rectum,  or  parameter  of  the  jjarabola. 
The  Latus-Rectum,  or  Parameter  of  the  parahola,  is  the 
double  ordinate  2)cissin(/  through  the  focus. 

The  abscissa  of  the  points  in  which  the  latus-rectum  pierces 

the  parabola  is  x  =-^-  . 

Making  this  substitution  in  the  equation 


if  =  2px 

we  have 

■    2/^  =  2p| 

=  F 

Hence 

2y=2p. 

CoR.  1.    Forming  a  proportion  from  the  equation 
2/2  =  2];jx,   . 
we  have  x  :  y  ::  y  :  1  p  ; 

i.e.,  the  latxis-rectum  of  the  jparahola  is  a  third  p>'^'02^ortional  to 
any  abscissa  and  its  corres^yonding  ordinate. 

EXAMPLES. 

Find  the  latus-rectum  and  write  the  equation  of  the  parab- 
ola which  contains  the  point : 

1.  (2,4).  3.    (a,b). 

Ans.     8,  y^  =  S  x.  Ans.     —,  y'^  =  —  x. 

a  a 

2.  (-2,4).  4.    {-a,  2). 

Ans.     —  8,  ?/2  =  —  8  X.  Ajis. ?  2/"  = ^• 

a  a 

5.    What  is  the  latus-rectnm  of   the  parabola  x'^^2py? 
How  is  it  defined  in  this  case  ? 


THE  PARABOLA.  87 

6.  What  is  the  equation  of  the  line  which  passes  through 
the  vertex  and  the  positive  extremity  of  the  latus-rectum  of 
any  parabola  whose  equation  is  of  the  form  ?/^  =  2  px  ? 

Ans.    y  =^  2  X. 

7.  The  focus  of  a  parabola  is  at  2  units'  distance  from  the 
vertex  of  the  curve  ;  what  is  its  equation 

(a)  when  symmetrical  with  respect  to  the  X-axis  ? 
\b)       "  "  "  "        "     "    Y-axis? 

Ans.    (a)  y^  =  8  cc,  (b)  cc^  =  8  y. 

Construct  each  of  the  following  parabolas  by  three  differ- 
ent methods. 

8.  ?/2  =  8  X.  10.   x^  =  6i/. 

9.  y"  =  —  4:  X.  11.    ic-  =  —  10  y. 

12.  What  are  the  co-ordinates  of  the  points  on  the  parabola 
y^  =  Q  X  where  the  ordinate  and  abscissa  are  equal  ? 

Ans.    (0,  0),  and  (6,  6). 

13.  Required  the  co-ordinates  of  the  point  on  the  parabola 
x^  —  4:y  whose  ordinate  and  abscissa  bear  to  each  other  the 
ration  3  :  2.  Ans.    (6,  9). 

14.  What  is  the  equation  of  the  parabola  when  referred  to 
the  directrix  and  X-axis  as  axes  ?  Ans.   y'^  =  2px  —  iJ^. 

Find  the  points  of  intersection  of  the  following : 

15.  y"  ^=  4x  and  2  y  —  cc  =  0. 

Ans.     (0,  0),  (16,  8). 

16.  03^  =  6  y  and  ?/  —  a;  —  1  =  0. 

17.  ?/2  =  —  8  a;  and  x  +  3  =  0. 

18.  ?/-  =  2  ic  and  x"  -\-  y"-  :=  8. 

Ans.    (2,  2),  (2,  -  2). 

19.  a;2  =  _  4  ?/  and  3  x^  +  2  y'-^  =  6. 

20.  x-  =  -iy  and  ?/-  =  4  x. 


88  PLANE  ANALYTIC   GEOMETRY. 

56.    To  deduce  the  polar  equation  of  the  ijarabola,  the  focus 
being  taken  as  the  pole. 

The  equation  of  the  parabola  referred  to  OY,  OX,  Fig.  29,  is 

y-  =  22:)x  ...   (1) 
To  refer  the  curve  to  the  initial  line  FX  and  the  pole  F 

-2 ,  0  )  we  have  for  the  equations  of  transformation,  Art.  34, 


Cor.  1, 


ic  =  -2  +  r  cos 


y  =  r  sm  6. 
Substituting  these  values  in  (1),  we  have 

r^  sin'  0  =^  p'  -\-  2  pr  cos  6. 
But  sin-  0  =  1  —  cos-  6  ; 

.-.  7-  :=  j;;2  _j_  2  237'  COS  6  +  '"^  cos^  ^  —  (^  -j-  /•  cos  Oy, 
,-.  r  ^  p  -\-  r  cos  0, 
or,  solving, 

r  =   -^ ...   (2) 

1  -  cos  ^  ^  ^ 

is  the  required  equation. 

We  might  have  deduced  this  value  directly  as  follows  : 
Let  P  (?',  &)  Fig.  29  be  any  point  on  the  curve  ;  then 

FP  =  DA  =  DF  +  FA  =p  +  r  cos  ^  ; 

i.  e.,  ?*  =  j9  -j-  r  cos  Q. 

Hence  r  = ^ . 

1  —  cos  0 

Cor.  1.    If     6  =  0,  r  =  co. 

li     e  =  90°,  r=p. 

If     6  =  180°,  r  =  I . 

If     ^  =  270°,  r  =  p. 
li     d  =  360°,  r  =  oo. 

An  inspection  of  the  figure  will  verify  these  results. 


THE  PARABOLA.  89 

57.  To  deduce  the  equation  of  the  tangent  to  the  •parabola. 
If  {x\  y'),  {x",  y")  be  the  points  in  which  a  secant  line  cuts 

the  parabola,  then 

y-y'=   y^-^^,  (x-x')    ...   (1) 

will  be  its  equation.     Since  (x',  y'),  (x",  y")  are  points  of  the 
parabola,  we  have 

y'^  =  2px'  ...   (2) 

y"^  =  2px"  ...   (3) 

These  three  equations  must  subsist  at  the  same  time ; 
hence,  subtracting  (3)  from  (2)  and  factoring,  we  have 

?/  -  y"  _     2^ 

i.e.,  x'  —  x"         y'  +  y" 

Substituting  this  value  in  (1),  the  equation  of  the  secant 
becomes 

y-y'  =  -^^  (x-x')  .  .  .  (4) 
y  +y 

When  the  secant,  revolved  about  (x",  y"),  becomes  tangent 
to  the  parabola  {x',  y')  coincides  with  {x",  y")  ;  hence  x'  =  x", 
y'  =  y".     Making  this  substitution  in  (4),  we  have, 

y  -  2/"  =  4  (^  -  ^") (5) 

y 

or,  simplifying,  recollecting  that  y""^  =  2  ^^x",  we  have 

^Jy"  =p(x  +  x")  ...   (6) 
for  the  equation  of  the  tangent  to  the  parabola. 

58.  To  deduce  the  value  of  the  sub-tangent. 
Making  ?/  =  0  in  (6),  Art.  57,  we  have 

x  =  -x"  =  OT,  (Fig.  31) 
for  the  abscissa  of  the  point  in  which  the  tangent  intersects 
the  X-axis.     But  the  sub-tangent  CT  is  the  distance  of  this 
point  from  the  foot  of  the  ordinate  of  the  point  of  tangency ; 
i.e.,  twice  the  distance  just  found;  hence 

Sub-tangent  =  2  x"  \, 


90 


PLANE  ANALYTIC  GEOMETRY. 


i.e.,  the  sub-tangent  is  equal  to  double  the  abscissa  of  the  point 
of  tangency. 

59.  The  preceding  principle  affords  us  a  simple  method  of 
constructing  a  tangent  to  a  parabola  at  a  given  point. 

Let  P"  (x",  y")  be  any  point  of  the  curve.  Draw  the  ordi- 
nate P"C,  and  measure  OC.     Lay  off  OT  =  OC. 


Fig.  31. 


A  line  joining  T  and  P"  will  be  tangent  to  the  parabola 
at  P". 

60.    To  deduce  the  equation  of  the  normal  to  the  parabola. 
The  equation  of  any  line  through  P"  (x",  y")  Fig.  31,  is 

y-y"  =  s(x-  x")  ...  (1) 
We  have  found  Art.  57,  (5)  for  the  slope  of  the  tangent  P"T 

hence,  for  the  slope  of  the  normal  P^IST,  v/e  have 
p  ' 


THE  PARABOLA.  91 

Substituting  this  value  of  s  in  (1),  we  have 

y  -y"  =  -  ^  (X  -  X")  ...  (2) 

for  the  equation  of  the  normal  to  the  parabola. 

61.  To  deduce  the  value  of  the  sub-normal. 

Making  y  =  0  in  (2)  Art.  60,  we  have,  after  reduction, 

x=p-\-x"  =  01^;  Eig.  31, 

.-.  Sid)-normal  =  ISTC  =p  -]-  a?"  —  x''  =  p. 

Hence  the  sub-normal  in  the  parabola  is  constant  and  equal 
to  the  semirparameter  FB. 

62.  To  show  that  the  tangents  drawn  at  the  extremities  of 
the  latiis  rectum,  are  perpendicular  to  each  other. 

The  co-ordinates  of  the  extremities  of  the  latus-rectum  are 

~ ,  p\  for  the  upper  point,  and  [-^  ,— p  j  for  the  lower  point. 

Substituting  these  values  successively  in  the  general  equa- 
tion of  the  tangent  line.  Art.  57  (6),  we  have 

yp  =p{x  +1 


or,  cancelling, 

y  =  x-\-P...(l) 

y=-x-^  .  .  .  (2) 

for  the  equations  of  the  tangents.  As  the  coefficient  of  x 
in  (2)  is  minus  the  reciprocal  of  the  coefficient  of  x  in  (1),  the 
lines  are  perpendicular  to  each  other. 

Cor.  1.    Making  ?/  =  0  in  (1)  and  (2),  we  find  in  each  case 

that  X  ^  —  ^ ;  hence,  the  tangents  at  the  extrem,ities  of  the 


-yp=p{x+l 


92 


PLANE  ANALYTIC   GEOMETRY. 


latus-rectum  and  the  directrix  meet  the  axis  of  the  parabola 
in  the  same  point. 

The  values  of  the  coefficients  of  x  in  (1)  and  (2)  show  that 
these  tangent  lines  make  angles  of  45°  with  the  X-axis. 

63.    To  deduce  the  equation  of  the  parabola  when  referred  to 
the  tangents  at  the  extremities  of  the  latus-rectum  as  axes. 


Fig.  32. 

The  equation  of  the  parabola  when  referred  to  OY,  OX,  is 

2/2  =  2px  .  .  .   (1). 

We  wish  to  ascertain  what  this  equation  becomes  when  the 
curve  is  referred  to  DY',  DX',  as  axes. 

Let  P'  (x',  y')  be  any  point  of  the  curve ;  then,  Fig.  32 

(OC,  CP')  =  (x,  y),  and  (DC,  C'P')  =  {x',  y'). 

Prom  the  figure,  we  have, 

OC  --  DC  -  DO  =  DK  +  CM  -  DO ; 


THE  PARABOLA.  93 


but 

DK  =  x'gos  45°  =  -^,  CM  =  /cos  45°  =^,  DO  =| 

V2  V2  2 


mce 

X 

X 

V2 

V2 

V 

2 

We 

have, 

also, 
11 

CP'  = 

_   y' 

=  MP'- 

x' 

C'K; 

I.e., 

V2         V2 

Substituting  the  values  of  x  and  y  in  (1),  we  have, 

\{:U'  -x'Y  ^^{x'  ^y')  -p'^  .  .  .   (2) 

In  order  to  simplify  this  expression  let  DP  =  a ;  then  from 
the  triangle  DPF,  we  have, 

DF  =  j9  =  a  cos  45°  =  -^ . 

V2i 

Substituting  this  value  of  j)  in  (2)  and  multiplying  through 
by  2,  we  have,    {y'  —  x'Y  =  2a  (x'  -\-  y')  —  a-, 
or,  tj'^  +  x'-^  -  2  x'y'  -  2  ax'  -2aij'  +  a^  =  0. 

Adding  4  x'y'  to  both  members,  the  equation  takes  the  form 
(x'  +y'  -ay  =  4.x'y', 
or  x'  -\-  y'  —  a  ^  ^2  a;'^  t/''^  ; 

.-.  transposing,  x'  -}^2  x'^ y'^  -\-  y'  ^  a; 

.:  x'^  ^y'}^=  ^a^-,  .  .  .  (3) 

or,  symmetrically,  dropping  accents, 

^;±2^=±l...(4) 
a^       a^ 

is  the  required  equation. 


94  PLANE  ANALYTIC  GEOMETRY. 

EXAMPLES. 

1.  What  is  the  polar  equation  of  the  parabola,  the  pole 
being  taken  at  the  vertex  of  the  curve  ? 

A71S.   r  ^=  2i  2)  cot  6  cosec  6. 

Find  the  equation  of  the  tangent  to  each  of  the  following 
parabolas,  and  give  the  value  of  the  subtangent  in  each  case : 

2.  7/2  =  4 £c  at  (1,  2).  Ans.   y  =  x  +  l;  2. 

3.  cc^  =  4  ?/  at  (—  2,  1).  Ans.    x  +  y  -i-l=0;  2. 

4.  y^  =  —  6xat  (—  6,  ord  +).  Ans.   2y  -\-x  =  6',  12. 

5.  x^  =  —  8  y  a.t  (abs  -{-,  —  2).  Ans.   x  -\-  y  =  2;  4. 

6.  y^  =  4  ax  at  (a,  —  2  a). 

7.  7/2  =  7JIX  at  (m,  ?7i). 

8.  iC'  =  —2^1/  ^t  (^bs  +,  — p). 


2py  at   ^a^-s  —  ,  I  j 


Write  the  equation  of  the  normal  to  each  of  the  following 
parabolas : 

10.  To  7/2  =  16  ic  at  (1,  4). 

11.  To  x'  =  —  lOy  at  (abs  +,  -  2). 

12.  To  7/2  =  —  mx  at  (—  m,  vi). 


13. 


To  ic2  =  2  7n,y  at  (  a^s  —  ,  —  ] 


14.  The  equation  of  a  parabola  is  x^  J-  y^  =  J-  a^ ;   what 
are  the  co-ordinates  of  the  vertex  of  the  curve  ? 

Ans.     na^-a 

\4:       '4 

15.  Given  the   parabola  y^  =  4x  and  the  line  y  —  x  =  0  ; 
required  the  equation  of  the  tangent  which  is, 

(a)  parallel  to  the  line, 

(b)  perpendicular  to  the  line. 

Ans.    (a)  y  =  x  -\-  1,  (b)  y  -\-  x  -\-  1  =  0. 


THE  PARABOLA.  95 

16.  The  point  (—1,  2)  lies  outside  the  parabola  y~  =  (Sx; 
what  are  the  equations  of  the  tangents  through  the  point  to 
the  parabola  ? 

17.  The  point  (2,  45°)  is  on  a  parabola  which  is  symmetri- 
cal with  respect  to  the  X-axis ;  required  the  equation  of  the 
parabola,  the  pole  being  at  the  focus. 

A71S.   /  =  (4  -  2  V2)  X. 

18.  The  subtangent  of  a  parabola  =  10  for  the  point  (5,  4)  ; 
required  the  equation  of  the  curve  and  the  value  of  the  sub- 
normal. 

Ans.     tf  =  —x\  p  . 

64.  The  tangent  to  the  parabola  makes  equal  angles  with  the 
focal  line  drawn  to  the  j^oint  of  tangency  and  the  axis  of  the 
curve. 

From  Fig.  31  we  have, 


FT  =  FO  -f  OT  =  £  -f  x' 

2^ 


We  have,  also, 


P 


FP"  =  DC  =  DO  +  OC  =  ^  +  x". 

.-.  FT  =  FP". 

The  triangle  FP"T  is  therefore  isosceles  and 
jrp"T  =  FTP". 

65.  To  find  the  condition  that  the  line  y  =  sx  -{-  c  must  fulfil 
in  order  to  touch  the  parabola  y^  =  2  px. 

Eliminating  y  from  the  two  equations,  and  solving  the 
resulting  equation  with  respect  to  x,  we  have, 


p  —  sc  ^^-\/  {cs  —  pY  —  cV 


•  •   (1) 


for  the  abscissae  of  the  points  of  intersection  of  the  parabola 
and  line,  considered  as  a  secant.      When  the  secant  becomes 


96  PLANE  ANALYTIC  GEOMETRY. 

a  tangent,  these  abscissas  become  equal ;  but  the  condition  for 
equality  of  abscissas  is  that  the  radical  in  the  numerator  of 
(1)  shall  be  zero  ;  hence 

(cs  —  p)-  —  c^s^  =  0, 

or,  solving  c  =  -^ 

""  2  s 

is  the  condition  that  the  line  must  fulfil  in  order  to  touch  the 

parabola. 

Cor.  1.    Substituting  the  value  of  c  in  the  equation 

y  =  sx  -\-  c, 

we  have,  y  =  sx  -\-  — —  ...  (2) 

^  s 

for  the  equation  of  the  tangent  in  terms  of  its  slope. 

66.  To  find  the  locus  generated  by  the  intersection  of  a  tan- 
gent, and  a  perpendicular  to  it  from  the  focus  as  the  point  of 
tangency  moves  around  the  curve. 

The  equation  of  a  straight  line  through  the  focus  | -^,  0  ]  is 

y=s'(x-P\...{l) 


In  order  that  this  line  shall  be  perpendicular  to  the  tangent 

y  =  sx  -\-  ^  .  .  .  (2) 
^  2s      ^  ^ 

we  must  have,    s'  =  —  -  ; 
s 

hence  y  =  —  -  a;  +  -^  ...   (3) 

s  2  s 

is  the  equation  of  a  line  through  the  focus  perpendicular  to 
the  tangent.     Subtracting  (3)  from  (2),  we  have 

s  -\ —  )  cc  ^  0, 

or,  ic  =  0, 

for  the  equation  of   the  required  locus.     But  cc  =  0  is  the 
equation  of   the  Y-axis ;    hence,  the  perjyendiculars  from  the 


THE  PARABOLA.  97 

focus  to  the  tangents  of  a  parabola  intersect  the  tangents  on  the 
Y-axis. 

67.  To  find  the  locus  genei'ated  by  the  inter'section  of  ttvo  tan- 
gents which  are. 'perpendicular  to  each  other  as  thepoints  of  tan- 
gency  moves  around  the  curve. 

The  equation  of  a  tangent  to  the  parabola  is,  Art.  65  (2), 

y  =  5^  +  -^  •  •  •  (1) 
The  equation  of  a  perpendicular  tangent  is 

y  =  _l^_^.     .     .     (2) 

Subtracting  (2)  from  (1),  we  have, 


S )  \  S  I  \L 

x  =  -l..  .  (3) 


is  the  equation  of  the  required  locus.  But  (3)  is  the  equa- 
tion of  the  directrix;  hence,  the  intersection  of  all  pjerpendicu- 
lar  tangents  drawn  to  the  parabola  are  points  of  the  directrix. 

68.  Two  tangents  are  drawn  to  the  parabola  from  a  point 
without ;  required  the  equation  of  the  line  joining  the  points  of 
tangency. 

Let  (x',  y')  be  the  given  point  without  the  parabola,  and  let 
(x",  y"),  (x2,  2/2)  be  the  points  of  tangency.     Since  (x',  ?/)  is 
on  both  tangents,  its  co-ordinates  must  satisfy  their  equations  ; 
hence,  the  equations  of  condition, 
y'y"=p(x'  +  x''), 

y%  =v  ix'  ^xo^. 

The  two  points  of  tangency  {x" ,  if),  (xo,  y^  must  therefore 
satisfy 

y'y  =p  (x'  -\-x'), 

or  yif  =p  {x  +x')  .  .  .   (1) 

Since  (1)  is  the  equation  of  a  straight  line,  and  is  satisfied 
for  the  co-ordinates  of  both  points  of  tangency,  it  is  the 
equation  of  the  line  joining  those  points. 


98 


PLANE  ANALYTIC  GEOMETRY. 


69.  To  find  the  equation  of  the  polar  of  the  pole  (x,'  y')  with 
regard  to  the  parabola  ^/^  =  2  px. 

The  pjolar  of  a  pole  with  regard  to  a  given  curve  is  the  line 
generated  by  the  point  of  intersection  of  a  pjair  of  tangents 
drawn  to  the  curve  at  the  p)oints  in  which  a  secant  line  through 
the  pjole  intersects  the  curve  as  the  secant  line  revolves  about  the 
pole. 

By  a  course  of  reasoning  similar  to  that  of  Art.  49,  we  may 
prove  tlie  required  equation  to  be 

yy'  =p  {x  ^x')   .  .  .   (1) 

As  the  reasoning  by  means  of  whicli  (1)  is  deduced  is  per- 
fectly general,  the  pole  may  be  without,  on,  or  within  the 
parabola. 

Cor.  1,    If  we  make,  in  (1),  {x' ,  ?/')  =  |  ^ ,  0  j,  we  have 

—  _-2  • 

^  —       2 ' 

hence,  the  directrix  is  the  p>olar  of  the  focus. 

70.  To  ascertain  the  position  and  direction  of  the  axes, 
other  than  the  axis  of  the  parabola  and  the  tangent  at  the 
vertex,  to  which  if  the  piarabola  be  referred  its  equation  will 
remain  unchanged  in  form. 


Fig.  33. 


THE  PARABOLA.  99 

Since  the  equation  is  to  retain  the  form 

•?/2  _  2pcc  .  .   .    (1) 
let  y'-'  =  2p'x'  ...  (2) 

be  the  equation  of  the  parabola  when  referred  to  the  axes, 
whose  position  and  direction  we  are  now  seeking.  It  is 
obvious  at  the  outset  that  whatever  may  be  the  position  of 
the  axes  relatively  to  each  other,  the  new  Y'-axis  must  be 
tangent  to  the  curve,  and  the  new  origin  must  be  07i  the 
curve ;  for,  if  in  (2)  we  make  x'  =  0,  Ave  have  ?/'  =  -|-  0,  a 
result  whicli  we  can  only  account  for  by  assuming  the  Y'-axis 
and  the  new  origin  in  the  positions  indicated.  This  conclu- 
sion, we  shall  see,  is  fully  verified  by  the  analysis  which 
follows. 

Let  us  refer  the  curve  to  a  pair  of  oblique  axes,  making 
any  angle  with  each  other,  the  origin  being  anywhere  in  the 
plane  of  the  curve.  The  equations  of  transformation  are. 
Art.  33  (1), 

X  =  a  -\-x^  cos  9  -{-  ■}/  cos  go 

y  =  b  -\-  x^  sin  0  -{-  y'  sin  go. 

Substituting  these  values  in  (1),  we  have, 

y'^  sin"^  <jD  +  2  x'y'  sin  0  sin  qd  +  x''^  sin^  $-\-2  (b  sin  cp  —  p 
cos  (f)  y'  +  2  (b  sin  0  —  p  cos  0)  x'  -{-  b"  —  2  pa  =  0  .  .  .   (3) 

Now,  in  order  that  this  equation  shall  reduce  to  the  same 
form  as  (1),  we  must  have  the  following  conditions  satisfied  : 

(a)  sin  0  sin  go  =  0. 

(b)  sin-^  ^  =  0. 

(c)  b^-  —  22)a  =  0. 

(d)  b  sin  (p  —  p  cos  (p  =  0. 

If  ^  =  0,  then  sin  9  sin  go  =  0  and  sin^  ^  =  0 ;  i.e.,  conditions 
(a)  and  (h)  are  satisfied  for  this  assumed  value  of  9.  But  9  is 
the  angle  which  the  new  X'-axis  makes  with  the  old  X-axis ; 
hence,  these  axes  are  parallel. 

If  (a,  h)  be  a  point  of  the  parabola  ?/  =  2  jyx,  then  b'^  =  2 
pa  is  an  analytical  expression  of  the  fact ;  hence  (c)  shows 
that  the  new  origin  lies  on  the  curve. 


100  PLANE  ANALYTIC  GEOMETRY. 

If      ^H_^  =  tan  g)  =  P,  then  (d)  is  satisfied.     But  ^  is  the 
cos  (p  b  0 

slope  of  the  tangent  at  the  point  whose  ordinate  is  b,  Art. 

57,  (5),  and  tan  cp  is  the  slope  of  the  new  Y'-axis ;  hence,  the 

new  Y'-axis  is  a  tangent  to  the  parabola  at  the  point  whose 

ordinate  is  6  ;  .-.  at  (a,  b)  ;  .:  at  the  new  origin. 

Cor.  1.    Substituting  (a),  (b),  (c),  and  (d)  in  (3),  recollecting 

that  cos  6  =  cos  0  =  1,  we  have,  after  dropping  accents, 

2/'  =  — ^  ^j 
sni-^  (p 

or,  letting  ■  /  ,     =  p', 

sm"^  qp 

we  have  y^  =  2p'x  ...  (4) 

for  the  equation  of  the  parabola  when  referred  to  O'Y',  O'X', 
Fig.  33.  -  The  form  of  (4)  shows  that  for  every  value  assumed 
for  X,  y  has  two  values,  equal  but  of  opposite  sign ;  hence, 
OX'  bisects  all  chords,  draw7i  parallel  to  0  Y'  and  is  therefore  a 
diameter  of  the  parabola. 

]SJ"oTE. A  Diameter  of  a  curve  is  a  line  which  bisects  a  sys- 
tem of  parallel  chords. 

71.  To  show  that  the  parameter  of  any  diameter  is  equal 
to  four  thnes  the  distance  from  the  focus  to  the  point  in  tvhich 
that  diameter  cuts  the  curve. 

Draw  the  focal  line  EO'  and  the  normal  O'N,  Fig.  33. 

Since  the  triangle   O'FT   is    isosceles.  Art.   64,  the  angle 

O'FN  =  2  (T. 

Since  O'N  is  a  normal  at  0',  AO'N  =  (f>  and  AN  =  p,  Art. 
61.     Hence  in  the  triangle  FO'A 

AO'  =  FO'  sin  2cp  =  FO'  2  sin  (jp  cos  q>. 

In  the  triangle  NO' A, 

,  ^,        .  T,x      ,  cos  (p 

AO  =  AN  cot  cp  =p  -^ — — ; 
sm  cp 

cos    QP 

hence  FO'  2  sin  go  cos  qp  =  ^^  —. ; 

sm  qp 


THE  PARABOLA. 


101 


But 


,.  YO'  =  — ^ — 
2/  =  ^J^-- 


Sin.-'  qp 
.-.  2/  =  4  FO'. 
72.    To  ^wcZ  ^7ie  equatiori  of  any  diameter  in  terms  of  the 
slope  of  the  tangent  and  the  semi-parameter. 

The  equation  of  any  diameter  as  O'X',  Fig.  33,  is 

y  =  AO'=  b. 
But  from  the  triangle  AO'N,  we  have, 

b  =  A^GOtq)  =  -^—=^', 

tan  (jp       s 


hence  y  =  -  ■  ■  ■  Q-) 

s 

is  the  required  equation. 

73.    To  show  that  the  tangents  draivn  at  the  extremities  of 
any  chord  meet  in  the  diameter  which  bisects  that  chord. 

p: 


R 

Y 

J^ 

^ 

M^ 

^ 

o'X^ 

/ 

\ 

1/ 

X 

0  f// 

/ 

y 

\ 

\ 

X' 


Fig.  34. 


Let  P'  {x',  y'),  P"  (x",  y")  be  the  extremities  of  the  chord 
then 


y-y'  =  l^i---^  ■■■(>-> 


102  PLANE  ANALYTIC   GEOMETRY. 

is  its  equation.     The  equation  of  the  tangents  at  P'  (x',  y'), 
P"  (x",  y")  are 

yy'  =- x>  (x  -\- x')  .  .  .  (2) 

yy"  =p  (x  +  x")  ...  (3) 

Eliminating  x  from  (2)  and  (3)  by  subtraction,  we  have, 

for  tlie  ordinate  of  tlie  point  of  intersection  of  the  tangents. 

x'  —  x" 
But is  the  reciprocal  of  the  slope  of    chord  P'P", 

y'  -  y" 

(see  (1)  ).     Hence,  since  the  chord  PT"  and  the  tangent  Y'T 
are  parallel,  we  have, 

x'  -x"  ^  1 

y'-y"~~s' 

Substituting  in  (4)  it  becomes 

V 
y  =  -• 
s 

Comparing    this    value    of   y   with    (1)    of   the   preceding 

article,  we  see  that  the  point  of  intersection  is  on  the  diameter. 


EXAMPLES. 

1.  What  must  be  the  value   of  c   in  order   that  the    line 
y  ^  4zX  -{-  c  may  touch  the  parabola  ?/^  =  8 cc  ? 

Ans.     \. 

2.  What  is  the  parameter  of  the  parabola  which  the  line 
y  =  3x  -\-2  touches  ? 

Ans.     24. 

3.  The  slope  of  a  tangent  to  the  parabola  y"^  =  Q  x  \&  =  3. 
What  is  the  equation  of  the  tangent  ? 

Ans.    y  =  3  0?  +  i- 

4.  The  point  (1,  3)  lies  on  a  tangent  to  a  parabola  ;  required 

the  equation  of  the  tangent  and  the  equation  of  the  parabola, 

the  slope  of  the  tangent  =  4. 

Ans.    y  =  4:X  —  1;  y^  =  —  16  x. 


THE  PARABOLA.  103 

5.  In  the  parabola  y-  =  8  a;  what  is  the  parameter  of  the 
diameter  whose  equation  is  ?/  —  16  =  0  ? 

Ans.    136. 

6.  Show  that  if  two  tangents  are  drawn  to  the  parabola 
from  any  point  of  the  directrix  they  will  meet  at  right  angles. 

7.  From  the  point  (—2,  5)  tangents  are  drawn  to  ?/^  =  8  a; ; 
required  the  equation  of  the  chord  joining  the  points  of 
tangency.  Ans.    5y  —  4a;  +  8  =  0. 

8.  What  are  the  equations  of  the  tangents  to  y-  =  Qx 
which  pass  through  the  point  (—  2,  4)  ? 

Find  the  equation  of  the  polar  of  the  pole  in  each  of  the 
following  cases : 

9.  Of  (-  1,  3)  with  regard  to  y"-  =  ^x. 

Ans.    3?/  —  2a;+2  =  0. 

10.  Of  (2,  2)  with  regard  to  i/-  =  -  4  a;. 

Ans.   2?/  +  2x  +  4  =  0. 

11.  Of  («,  h)  with  regard  to  ?/^  =  4  x. 

Ans.    by  —  2  x  —  2  a  ^  0. 

12.  Given  the  parabola  y"^  ^  x  and  the  point  (—  4,  10)  ;  to 
find  the  intercej)ts  of  the  polar  of  the  point. 

Ans.     a  =  4,  b  = . 

5 

13.  The  latus-rectum  of  a  parabola  =  4 ;  required  the  pole 
of  the  line  y  —  8a:;  —  4  =  0. 

Ans.     (1   1). 

14.  Given  y^  =  10  x  and  the  tangent  2y  —  x  =  10;  required 
the  equation  of  the  diameter  passing  through  the  point  of 
tangency. 

Ans.     y  =  10. 

GENERAL  EXAMPLES. 

1.  Assuming  the  equation  of  the  parabola,  prove  that  every 
point  on  the  curve  is  equally  distant  from  the  focus  and 
directrix. 


104  PLANE  ANALYTIC  GEOMETRY. 

2.  Find  the  equation  of  the  parabola  which  contains  the 
points  (0,  0),  (2,  3),  (-  2,  3). 

Ans.     3x^  =  4:y. 

3.  What  are  the  parameters  of  the  parabolas  which  pass 
through  the  point  (3,  4)  ? 

A71S.     J/,  and  |. 

4.  Find  the  equation  of  that  tangent  to  y^  =  9  x  which  is 
parallel  to  the  line  y  —  2x  —  4z  =  0. 

Ans.     8y  —  16x  —  9  =  0. 

5.  The  parameter  of  a  parabola  is  4 ;  required  the  equation 
of  the  tangent  line  which  is  perpendicular  to  the  line 
y  =  2  X  -\-  2.  Give  also  the  equation  of  the  normal  which  is 
parallel  to  the  given  line. 

6.  A  tangent  to  y^  =  4:X  makes  an  angle  of  45°  with  the 

X-axis  ;  required  the  point  of  tangency. 

A71S.     (1,  2). 

Show  that  tangents  drawn  at  the  extremities  of  a  focal 
chord 

7.  Intersect  on  the  directrix. 

8.  Meet  at  right  angles. 

9.  That  a  line  joining  their  point  of  intersection  with  the 
focus  is  perpendicular  to  the  focal  chord. 

10.  Find  the  equation  of  the  normal  in  terms  of  its  slope. 

11.  Show  that  from  any  point  within  the  parabola  three 
normals  may  be  drawn  to  the  curve. 

12.  Given  the  parabola  r  = to  construct  the  tan- 

^  1  +  cos  ^ 

gent  at  the  point  whose  vectorial  angle  =  60°,  and  to  find  the 
angle  which  the  tangent  makes  with  the  initial  line. 

A71S.    6  =  60°. 

13.  Find  the  co-ordinates  of  the  pole,  the  normal  at  one 
extremity  of  the  latus-rectum  being  its  polar. 


THE  PARABOLA.  105 

14.  Ill  the  parabola  y^  =  4  ck  what  is  the  equation  of  the 
chord  which  the  point  (2,  1)  bisects  ? 

Ans.     y  =  2  X  —  3. 

15.  The  polar  of  any  point  in  a  diameter  is  parallel  to  the 
ordinates  of  that  diameter. 

16.  The  equation  of  a  chord  of  y"^  =  10  x  \s  y  =  2  x  —  1; 
required  the  equation  of  the  corresponding  diameter. 

17.  Show  that  a  circle  described  on  a  focal  chord  of  the 
parabola  touches  the  directrix. 

18.  The  base  of  a  triangle  =  2  a  and  the  sum  of  the  tan- 
gents of  the  base  angles  =  b.  Show  that  the  locus  of  the 
vertex  is  a  parabola. 

19.  Required  the  equation  of  the  chord  of  the  parabola 
y^  =:2px  whose  middle  point  is  (m,  n). 

.  n        X  —  m 

Ans.     —  = . 

p         y  —  n 

20.  A  focal  chord  of  the  parabola  y"^  =  2px  makes  an 
angle  =  <jo  with  the  X-axis ;  required  its  length. 

Ans.     —~—  • 
sin^  qi 

21.  Show  that  the  focal  distance  of  the  point  of  intersec- 
tion of  two  tangents  to  a  parabola  is  a  mean  proportional  to 
the  focal  radii  of  the  points  of  tangency. 

22.  Show  that  the  angle  between  two  tangents  to  a  parab- 
ola is  one-half  the  angle  between  the  focal  radii  of  the  points 
of  tangency. 

23.  The  equation  of  a  diameter  of  the  parabola  y^  =  2px 
is  y  =  a;  required  the  equation  of  the  focal  chord  which  this 
diameter  bisects. 

24.  The  polars  of  all  points  on  the  latus-rectum  meet  the 
axis  of  the  parabola  y^  =  2px  in  the  same  point ;  required  the 
co-ordinates  of  the  point. 

Ans.     (-^,0 


106 


PLANE  ANALYTIC   GEOMETRY.. 


CHAPTER    VII. 
THE   ELLIPSE. 

74.  The  ellipse  is  the  locus  of  a  point  so  moving  in  a  plane 
that  the  sum  of  its  distances  from  two  fixed  points  is  always 
constant  and  equal  to  a  given  line.  The  fixed  points  are 
called  the  Foci  of  the  ellipse.  If  the  points  are  on  the 
given  line  and  equidistant  from  its  extremities,  then  the  given 
line  is  called  the  Transverse  or  Major  Axis  of  the  ellipse. 

75.  To  deduce  the  equation  of  the  ellii^se,  given  the  foci  and 
the  transverse  axis. 

Y 


Fig.  35. 


Let  F,  Fi  be  the  foci  and  AA'  the  transverse  axis.  Draw 
OY  J_  to  AA'  at  its  middle  point,  and  take  OY,  OX  as  the 
co-ordinate  axes. 


THE  ELLIPSE.  107 

Let  P  be  ar.y  point  of  the  curve.  Draw  PF,  PFj ;  draw 
also  PD  II  to  OY. 

Then  (OD,  DP)  =  (x,  y)  are  the  co-ordinates  of  P. 

Let  AA'=  2a,  FFi  =  20F  =  20Fi  =2  c,  FP  =  r  and 
FiP  =  /. 

From  the  ric^ht  angled  triangles  FPD  and  FjPD,  we  have, 


r  =  ^if  +  (cc  —  cf  and  r'  =  Vy'  +  (a?  +  c)^  ,  .   .   (a) 
From  the  mode  of  generation  of  the  curve,  we  have, 
?•  +  ?•'  =  2  a  ; 


hence  Vy'  +  (a;  -  c)^  +  V  ^f+  (cc  +  c)^  =  2  a  ;  .  .  .   (1) 

or,  clearing  of  radicals,  and  reducing, 

a^  (y-2  +  a;^)  -c'x'  =  a'  (a^  -  c')  ...   (2) 

As  this  equation  (2)  expresses  the  relationship  between  the 
co-ordinates  of  any  point  on  the  curve,  it  must  express  the 
relationship  between  the  co-ordinates  of  every  point ;  hence 
it  is  the  required  equation. 

Equation    (2)  may   be   made,   however,  to  assume  a  more 
elegant  form.     Make  cc  =  0  in  (2),  we  have, 
y"^  z=  a^  —  c^ 

for  the  square  of  the  ordinate  of  the  point  in  which  the 
curve  cuts  the  Y-axis;  i.e.,  OB^  (=  OB'^).  Eepresenting 
this  distance  by  b,  we  have, 

b^  =  a^  -  c2, 

.-.  c^  =  a^  -b^    ...   (3) 

Substituting  this  value  of  c^  in  (2)  and  reducing,  we  have, 
a^  if  +  b^  x^  =  aH- ;  .  .  .  (4) 
or,  symmetrically, 

£!  +  ^  =  l...(5) 
a''        b^ 

for  the  equation  of  ellipse  when  referred  to  its  centre  and 
axes. 

Let  the  student  discuss  equation  (4).     See  Art.  12. 


108  PLANE  ANALYTIC  GEOMETRY. 

Cor.  1.    If  we  make  h  =  a  in  (4),  we  have, 
x^  -\-  y"^  =  a^ 

whicli  is  the  equation  of  a  circle. 

Cor.  2.    If  we  interchange  a  and  b  in  (5),  we  have, 

■#  +  -4  =  1  •  •  •  (6) 

for  the  equation  of  an  ellipse  whose  transverse  axis  (=  2  a) 
lies  along  the  Y-axis. 

CoR.  3.  If  {x',  y')  and  (x",  y")  are  two  points  on  the  curve, 
we  have  from  (4) 

3/'2  =  i!  (a2  _  cc'2)  and  y"^  =^^{a}  -  x'"")  ; 

hence,  y'^  :  t/"^  v.  (a  -  x')  {a  +  x')  :  {a  —  x")  {a  +  x")  ; 
i.e.,  the  squares    of  the   ordinates  of  any  two  points  on  the 
ellipse  are  to  each  other  as  the  rectangles  of  the  segments  in 
which  they  divide  the  transverse  axis. 

CoR.  4.  By  making  x  =  x'  —  a  and  y  =  y'  in  (4),  we  have 
after  reduction  and  dropping  accents, 

a^y^ -\-Px^ -2ab^x  =0  .  .   .   (7) 

for  the  equation  of  the  ellipse,  A'  being  taken  as  the  origin 
of  co-ordinates. 

76.  The  line  BB',  Fig.  35,  is  called  the  Conjugate  or 
Minor  axis  of  the  ellipse ;  the  points  A  and  A'  are  called  the 
Vertices  of  the  ellipse.  It  is  evident  from  the  figure  that 
the  point  0  bisects  all  lines  drawn  through  it  and  terminating 
in  the  curve.  For  this  reason  0  is  called  the  centre  of 
the  ellipse. 

The  ratio         vV_-_&2  ^  c  ^  ^^     g^^  .^.  ^^^_  75  .  .  .  (1) 
a  a 

is  called  the  Eccentricity  of  the  ellipse.  It  is  evident  that 
this  ratio  is  always  <  1.  The  value  of  c  =  J-  Va^  —  b^  meas- 
ures the  distances  of  the  foci  F,  Fi  from  the  centre. 


THE  ELLIPSE.  109 

li  a  =b  in  (1),  then  e  =  0 ;  i.e.,  when  the  ellipse  becomes  a 
circle  its  eccentricity  becomes  zero. 

If  Z»  =  0  in  (1),  then  e  =  1 ;  i.e.,  when  the  ellipse  becomes  a 
straight  line  the  eccentricity  becomes  unity. 

77.  To  find  the  values  of  the  focal  radii,  r,  r',  of  a  point  on 
the  ellipse  in  terms  of  the  abscissa  of  the  j^oint. 

The  Focal  Radius  of  a  point  on  the  ellipse  is  the  distance 
of  the  point  from  either  focus. 

From  equations  (a).  Art,  75,  we  have, 

r  =  ^  if  -\-{x  -cf; 
from  the  equation  of  the  ellipse,  Art.  75  (4),  we  have, 

7/2   =  ^  (a2   -  X')    =  ?»2   _  1:  ^2; 

cr  «" 

hence,  substituting 

V 


r  =  Jb''  -  -^  x2  4-  a;2  -  2  cic  +  c2 


=  Jc^j^b'~2cx^  "—^  x\ 


=  K    a"  -  2  ex  +  ^x"" 
V  a^ 


c 

=  a- X  \ 

a 

hence  r  =  a  —  ex.     See  (1)  Art.  76  .  .  .  (1) 

Similarly  we  find 

r'  =  a  -\-  ex  .  .  .   (2) 

78.  Having  given  the  transverse  axis  and  the  foci  of  any 
ellipse,  the  principles  of  Art.  75  enables  us  to  construct  the 
ellipse  by  three  different  methods. 

First  Method.  — Take  a  cord  equal  in  length  to  the  trans- 
verse axis  AA'.  Attach  one  end  of  it  at  F,  the  other  at  F'. 
Place  the  point  of  a  pencil  in  the  loop  formed  by  the  cord 
and  stretch  it  upward  until  taut.  Wheeling  the  pencil  around, 
while  keeping  the  point  on  the   paper  and  tightly  pressed 


110 


PLANE  ANALYTIC  GEOMETRY. 


against  the  cord,  the  path  described  will  be  an  arc  of  the 
ellipse.  After  describing  the  upper  half  of  the  ellipse,  re- 
move the  pencil  and  form  the  loop  below  the  transverse  axis. 
By  a  similar  process  the  lower  half  may  be  described.     It  is 


Fig.  36. 


evident  during  the  operation  that  the  sum  of  the  distances  of 
the  point  of  the  pencil  from  the  foci  is  constant  and  equal  to 
the  length  of  the  cord ;  i.e.,  to  the  transverse  axis. 

Second  Method.  —  Take  any  point  C  on  the  transverse  axis 
and  measure  the  distances  A'C,  AC.  With  F'  as  a  centre  and 
CA'  as  a  radius  describe  the  arc  of  a  circle  ;  also  with  F  as  a 
centre  and  CA  as  a  radius  describe  another  arc.  The  points 
E.,  R'  in  which  these  arcs  intersect  are  points  of  the  ellipse. 
By  interchanging  the  radii  two  other  points  P,  P'  may  be 
determined.  A  smooth  curve  traced  through  a  number  of 
points  thus  located  will  be  the  required  ellipse. 

Third  Method.  —  Let  the  axes  AA'  =  2  a,  BB'  =  2  &  be 
given.  Lay  off  on  any  straight  edge  MIST  (a  piece  of  paper 
will  do)  KD  =  OA  =  «  and  DL  =  OB  =  b.  Place  the 
straight  edge  on  the  axes  in  the  position  indicated  in  the 
figure.     Then  as  K  and  L  slide  along  the  axes,  the  point  D 


THE  ELLIPSE.  Ill 

will  describe  the  ellipse.  For  from  the  figure  DLH  and 
DKE  are  similar  triangles  : 

= ;  I.e.,  —  ^  — z===i  (x  and  y  being 

•••       ,    KE        LH '       '  cc       -^h^^if 

the  co-ordinates  of  D). 

Hence,  squaring,  clearing  of  fractions,  and  transposing,  we 
have 

a'^y'^  +  ^"^^  =  (i^V^- 

That  is  the  locus  described  by  D  is  an  ellipse.  An  instru- 
ment based  upon  this  principle  is  commonly  used  for  drawing 
the  ellipse. 

79.    To  find  the  latus  rectum,  or  parameter  of  an  ellipse. 

The  latus  rectum  or  parameter  of  an  eUi-pse  is  the  double 
ordinate  passing  through  the  focus. 

The  abscissas  of  the  points  in  which  the  latus  rectum 
pierces  the  ellipse  are  a;  =  -j-  Va^  —  h^.  Substituting  either 
of  these  values  on  the  equation  of  the  ellipse 

a^ 

we  have  y^  =  -—  (a"^  —  (or'  —  If))  =—-.•.?/  =  — . 

a"  a"  a 

2  If 
Hence  Latus  rectum  =  2  ?/  =  — —  ...  (1) 

a 

Forming  a  proportion  from  this  equation  there  results, 

2y:2b::h:a; 
^  hence  2y:2h  ■.■.2b -.2  a; 

i.e.,  the  latus  rectum  is  a  third  proportional  to  the  tivo  axes. 

EXAMPLES. 

Find  the  semi-axes,  the  eccentricity,  and  the  latus  rectum 
of  each  of  the  following  ellipses  : 

1.  3£c2^2y2  =  6.  3.    a;2^3,y-==2. 

2.  -^  +  i^  =  1.  4.    4  7/^  -f  6  =  8  -  2  xl 


7. 

f^ 

2    ^'^' 

8. 

^2  + 

^^  =  n. 

112  PLANE  ANALYTIC   GEOMETRY. 

6.    c?/^  -(-  cc-  =  c?. 

Write  the  equation  of  the  ellipse  having  given : 

9.  The  transverse  axis  =  10 ;  the  distance  between  the  foci 
=  8. 

Ans.     — — h  —  =  1- 

25  ^   9 

10.  Sum  of  the  axes  =  18  ;  difference  of  axes  =  6. 

Ans.     ^  +  i^  =  1. 
36         9 

11.  Transverse   axis  =  10 ;    the   conjugate   axis  =  ^    the 
transverse  axis. 

Ans. 1 =^  =  1. 

25         25 

12.  Transverse  axis  =  20 ;    conjugate   axis  =  distance  be- 
tween foci. 

Ans.     —  +  3/2  =  50. 

13.  Conjugate  axis  =  10  ;  distance  between  foci  =  10. 

Ans.     ~-\-  f  =  25. 

14.  Given  3  y^  -\-  4^  x^  =  12 ;  required  the  co-ordinates  of  the 
point  whose  ordinate  is  double  its  abscissa. 

VI'VI 


Ans. 


15.  Given  the  ellipse  oy'^  -\-2x^  =  12,  and  the  line  y  =  x  — 1; 
to  find  the  co-ordinates  of  their  points  of  intersection. 

16.  Given  the  ellipse  - — 1-^  =  1,  and   the  abscissa  of  a 

^      64       15         ' 

point  on  the  curve  =  ^;  required  the  focal  radii  of  the  point. 

Ans.     r  =  7j%  r'  =  S^^. 


THE  ELLIPSE. 


113 


80.    To  deduce  the  •polar  equation  of  the  ellipse,  either  focus 
being  taken  as  the  pole. 


Fig.  37. 

Let  us  take  F  as  the  pole,  and  let  (FP',  P'FA)  =  (r,  6)  be 
the  co-ordinates  of  any  point  P'  of  the  ellipse.  From  Art.  77 
(1)  we  have,      r  ^=  a  —  ex'  .  .  .  (V) 

From  the  figure,  OD  =  OF  +  FD ; 
i.e.,  x'  ^  ae  -\-  r  cos  6. 

Substituting  this  value  of  x'  in  (1),  we  have 

r  ^  a  —  e  {ae  -\-  r  cos  6), 

or,  reducing,  we  have 

,,  _   ^^  (1  -  ^')  ^2) 

1  +  e  cos  ^   ■  ■  ■   ^  ^ 

for  the  polar  equation   of   the  ellipse,  the  right-hand  focus 
being  taken  as  the  pole. 
From  Art.  77  (2), 

FT'  =  /  =  a  +  ex'. 

We  readily  determine  from  this  value 


^.  _   g  (1  -  e^) 
1  —  e  cos  6 


(3) 


for  the  polar  equation  of  the  ellipse,  the  left-hand  focus  being 
taken  as  the  pole. 


114 

Cor.  If 

If 


PLANE  ANALYTIC  GEOMETRY. 

e  =  0,         r  =  a  (1  -  e)  ='FA, 
r'  =  «  (1  +  e)  =  F'A. 

e  =  90°,        r  =  a{l-  e')=a  —  a 


cv"  -  b^ 


=  —  =  FM. 

a 


(1  —  e^)  =  a  —  a 


a^  —  6^ 


-  =  F'N. 
a 


If 
If 

If 


6  =  180°,    r  =  a  (1  +  e)  =  FA', 
/  =  «  (1  _  e)  =  F'A'. 

6  =  270°,    ?■  =  ft  (1  -  e^)  =  FM', 
r'  =  «  (1  -  e^)  =  F'N'. 

^  =  360°,    r  =  «  (1-  e)  =  FA, 
r'  =  a  (1  +  e)  =  F'A. 


81.    ^0  deduce  the  equation  of  condition  for  the  supplemental 
chords  of  an  ellipse. 


Fig.  38. 


Let  AP,  AT  be  a  pair  of  supplemental  chords. 
The  equation  of  a  line  through  A  (a,  a)  is 

y  =  s  (x  —  <x). 


THE  ELLIPSE.  115 

The  equation  of  a  line  through  A'  (—  a,  o)  is 

y  =  s'{x  +  a). 
Where  these  lines  intersect  we  must  have 

In  order  that  the  lines  shall  intersect  07i  the  ellipse  their 
equations  must  subsist  at  the  same  time  with  the  equation  of 
the  ellipse 

^/  =  l,{a'^-x')  ...   (2) 

Dividing  (1)  by  (2),  we  have 
1  = ss  •. 

or  ,/  =  _  il  .  .  .   (3) 

for  the  required  condition. 

CoR.  If  a  =  b,  the  ellipse  becomes  a  circle  and  (3)  be- 
comes 

ss' =    -1, 

a  relationship  heretofore  deduced.     Art.  40  (1). 

ScHOL.  The  preceding  discussions  have  developed  a  remark- 
able analogy  between  the  ellipse  and  circle.  As  we  proceed 
we  shall  find  that  the  circle  is  only  a  particular  form  of  the 
ellipse  and  that  all  of  the  equations  pertaining  to  it  may  be 
deduced  directly  from  the  corresponding  equations  deduced 
for  the  ellipse  by  simply  making  a  =  b  in  those  equations. 

82.    To  deduce  the  equation  of  the  tangent  to  the  ellipse. 
Let  P"  (x",  y"),  P'  (x',  y')  be  the  points  in  which  a  secant 
P''S  cuts  the  ellipse.     Its  equation  is,  therefore, 

y-y'  =  y^f^,{x-x')  ...  (1) 

As  the  points  are  on  the  ellipse,  we  must  have 

y"  =  ^{ci^-x'^)  ...  (2) 

2/"2  =  il /(j2  _  ^./m  ...  (3) 
a^ 


116 


PLANE  ANALYTIC  GEOMETRY. 


Fig.  39. 


These  three  equations  must  subsist  at  the  same  time ;  hence 
subtracting  (3)  from  (2)  and  factoring,  we  have 


(/  -  y")  {y'  +  y")  = 


(x'  -  x")  (x'  +  x")  ; 


hence 


y  -  If 


V"    x'  +  x" 


y'  +  /' 
Substituting  this  value  in  (1)  it  becomes 

b"^     x'  -\-  X 
a" 


y-y  ---IT 


(x  —  x'). 


y'  +  y" 

Eevolving  the  secant  line  upward  about  the  point  Y"  (x",  y") 
the  other  point  of  intersection  P'  (x',  y')  will  approach  P''  and 
will  finally  coincide  with  it.  When  this  occurs  the  secant 
becomes  a  tangent  and  cc'  =  x",  y'  =  y"  -^  hence,  substituting, 
we  have 

7,2         r^'l 

y  —  y  == ^•^7(^-»'); 

a^      y 
ahjy"  +  b^xx"  =  a'b'';  ...   (4) 


i.e 
or 


+ 


yy 


a-  b'^ 

for  the  equation  of  the  tangent. 


=  1  . 


(5) 


THE  ELLIPSE.  117 

CoK.    li  b  =  a,  we  have 

for  the  equation  of  the  tangent  to  the  circle.     See  Art.  41  (6). 
ScHOL.    If  we  make  x  and  y  successively  =  0  in  the  equa- 

tion  of  the  tangent  (5),  we  have  y  =  —,  and  cc  =  -^  for  the 

y  X 

values  of  the  variable  intercepts  OT,  OT,  Fig.  39 ; 

hence  'if'  =  —  and  x"  =  — . 

y  ^ 

These  values  in  the  equation 

a"         V" 
give,  after  reduction, 

^  +  4  =  1  .  .  •  (6) 
X-         2/2 

for  the  equation  of  the  ellipse,  the  intercepts  of  its  tangents 
on  the  axes  being  the  variables. 


83.    To  deduce  the  value  of  the  sub-tangent. 
Making  ?/  =  0  in  (5),  Art.  82,  we  have 


ic  =  OT  =  ^ 

x'' 


a^  —  x: 


,\  sub-tangent     =  DT  =  -^ x"  = 

x"  x 

CoK.  li  b  =  a,  then  from  Art.  41,  Schol.  a""  —  x"^  =  tj"% 
.'.  sub-tangent  in  the  circle  =  ^^  . 

X 

Schol.  The  value  of  the  sub-tangent  being  independent  of 
the  value  of  the  minor  axis  (2  b)  it  follows  that  this  value  is 
the  same  for  every  ellipse  which  is  concentric  with  the  given 
ellipse,  and  whose  common  transverse  axis  is  2  a. 


118  PLANE  ANALYTIC  GEOMETRY. 

84.  The  equation  of  condition  that  a  line  shall  pass  through 
the  centre  of  the  ellipse  and  the  point  of  tangency  is,  Fig.  39, 

.:  the  slope  of  this  line  is 

x" 

The  slope  of  the  tangent  at  (x",  y")  is.  Art.  82, 

t'  =-^    ^ 
o?  '  y"  ' 

Multiplying,  member  by  member,  we  have 

tt'=-^  ...  (1) 
But  Art.  81   (3) 

a' 

.-.  s/  =  tt' ; 

i.e.,  the  tangent  to  the  ellipse  and  the  line  joining  the  centre  and 
the  point  of  tangency  enjoy  the  property  of  being  supplemental 

chords  of  an  ellipse  whose  semi-axes  bear  to  each  other  the  ratio  — . 

CoR.  J.i  s  =  t,  then  s'  =  t' ;  i.e.,  if  one  supplementary  chord 
is  parallel  to  a  diameter  of  the  ellipse,  the  other  supplementary 
chord  is  parallel  to  the  tangent  dratvn  at  the  extremity  of  that 
diameter. 

85.  The  principles  of  Arts.  83,  84  afford  us  two  different 
methods  of  constructing  a  tangent  to  the  ellipse  at  a  given 
point. 

First  Method.  —  Art.  83,  Schol.  Let  V,  Fig.  40,  be  the  given 
point.  Through  P"  draw  the  ordinate  P''D  and  produce  it 
until  it  meets  the  circle  described  upon  the  transverse  axis 
of  the  ellipse  (A A')  in  P';  draw  P'T  tangent  to  the  circle 
at  P'.     Join  P"  and  T ;  P"T  will  be  the  required  tangent. 


THE  ELLIPSE. 


119 


Fig.  40. 


Second  Method.  —  Art.  84  and  Cor.  Draw  P"R  througli  the 
centre,  and  from  A'  draw  A'E'  ||  to  P"R ;  P"T  drawn  through 
V"  II  to  R'A  will  be  tangent  to  the  ellipse  at  P". 

86.    To  deduce  the  equation  of  the  normal  to  the  ellipse. 
The  equation  of  any  line  through  V  (x",  y"),  Pig.  39,  is 

^J  -  y"  =  s  (a;  -  x")  .  .  .   (1). 
In  order  that  this  line  and  the  tangent  at  P"  (x" ,  y")  shall 
be  perpendicular  their  slopes  must  satisfy  the  condition 
1  +  s/  =  0  .  .  .  (2). 

We  have  found  Art.  82  for  the  slope  of  the  tangent 
J  _       b''      x" . 


a- 


r 


hence,  the  slope  of  the  normal  is 
s 


*     y 


Substituting  this  value  of  s  in  (1),  we  have 
for  the  equatio7i  of  the  Jiormal  to  the  ellipse. 


120  PLANE  ANALYTIC  GEOMETRY. 

Cor.  1.    If  a  =  b,  then  (3)  becomes,  after  reduction, 

yx"  —  xy"  =  0, 
which  is  the  equation  of  the  normal  line  to  the  circle. 

87.    To  deduce  the  value  of  the  sub-normal. 

Making  y  =  0  in  the  equation  of  the  normal,  (3),  Art.  86, 

^2   /,2 

we  have,  Fig.  39,    ON  =  a;  = x"  ==  e^x" , 

a^ 

.-.  Sub-normal  =  jm=  x"  -  ^'  ~  ^'  x"=—  x" . 

a?  a^ 

Cor.  1.    li  a  =  b,  then 

Sub-normal  for  the  circle  =  x". 


EXAMPLES. 

1.    Deduce  the  polar  equation  of  the  ellipse,  the  pole  being 
at  the  centre  and  the  initial  line  coincident  with  the  X-axis. 

Ans.     r  =  — —  . 


V«'  sin2  ^  +  ^»2  cos^  e 

Write  the  equation  of  the  tangent  to  each  of  the  following 
ellipses,  and  give  the  value  of  the  sub-tangent  in  each  case. 

2.  2  a;2  +  4  2/2  =  38  at  (1,  3). 

Ans.     cc  +  6  ?/  =  19  ;  18. 

2  2 

3.  -^ — ^  ^  ^  1^     at  (1,  ordinate  positive). 

o  ^ 

3 
Ans.     X  -\-  ~^v  =  3  :    2. 
WB 

4-    ^  +  ^  =  1,     at  (2,0). 

Ans.     X  =  2;  0. 

5.  2  a;2  _^  3  7/2  =  11  at  (2,  -  1). 

Ans.     4:X  —  3y  =  11;  |. 

6.  J^  +  ^  =  1,     at  (0,  V«)  . 


THE  ELLIPSE.  121 


7.    ^  +  -1^=1,     at(a,o) 


8.  i/  +  hx""  -=  2,     at  (1,  -  V2  -  Z*). 

9.  — \-  if-  =  1,     at  (abs  -\-,  .5). 

Write  the  equation  of  the  normal  to  each  of  the  following 
ellipses,  and  give  the  value  of  the  sub-normal. 

10.  3  ?/2  +  4  a;2  =  39,    at   (3,  1). 

11.  4  2/-  +  2  a;2  =  44,    at    (—  2,  ord  negative). 

12.  -^  +  1^  =  1,     at  (-1,  ord-). 

13.  ^  +  1^  =  1,     at  (1,  2). 

14.  ^  +  f  =  l,     at(i,ord+). 

a 

15.  vi^y'^  -\-  iv'x^  =  mhi-,     at  (m,  o). 

16.  The  equation  of  a  chord  of  an  ellipse  is?/^  —  2a;  +  6; 

what  is  the  equation  of  the  supplementary  chord,  the  axes  of 

the  ellipse  being  6  and  4  ? 

Ans.     ?/  =  I  a;  +  f . 

2  2 

17.  Given  the  equation  — — f-  -^  =  1,  and  ?/  —  2  =  0 ;  re- 
quired the  equation  of  the  tangents  to  the  ellipse  at  the  points 
in  which  the  line  cuts  tne  curve. 

18.  Given  the  ellipse  - — \-  ^  =  1,  and  the  line  y  —  x  -\- 

4  9 

2  =  0;  required 

(a)  The  equation  of  a  tangent  to  the  ellipse  ||  to  the  line. 

/^^      ci  u  u  u  (.(,  u        J_  a 


a       a 


122 


PLANE  ANALYTIC  GEOMETRY. 


19.    The  point  (4,  3)  is  outside  the  ellipse 

16  ^  9  ~     ' 

required  the  equations  of  the  tangents  to  the  ellipse  which 
pass  through  the  point. 

88.    The  angle  formed  hy  the  focal  lines  drawn  to  any  point 
of  an  ellipse  is  bisected  by  the  normal  at  that  point. 


Fig.  41. 

Let  F'N  be  a  normal  at  any  point  P"  (x",  y").    Draw  P"F,  P"Fo' 
We  have  found,  Art.  87,  that 

^2   }fi 

From  Art.  76  we  have  OF  =  OF'  =  ae ;  hence 

NF  =  OF  -  ON  =  ae  -  eV  =  e  (a  -  ex'') 
NF'  =  OF'  +  ON  =  ae  +  e'x"  =  e  (a -\-  ex") 
.'.    NF  :  NF'  ::  (a-  ex")  :  (a  +  ex") 

But  FP"  :  F'P"  ::  (a-  ex")  :  (a  +  ex")  Art.  77,  (1)  and  (2)5 
.'.    NF  :  NF'  ::  FP"  :  F'P". 


THE  ELLIPSE.  123 

The  normal,  therefore,  divides  the  base  of  the  triangle 
F'P"F  into  two  segments  which  are  proportional  to  the  adja- 
cent sides.     Hence 

FP"N  =  FT"N. 
ScHOL.  1.    If  P"T  be  a  tangent  drawn  at  V",  we  must  have 
FT''C  =  FP"T ; 

for  each  of  these  angles  is  equal  to  the  difference  between  a 
right  angle  and  the  angle  FT"N  (=  FP"N).  Hence,  the 
tangent  to  the  ellipse  makes  equal  angles  with  the  focal  radii 
drawn  to  the  point  of  tangency. 

ScHOL.  2.  The  principles  of  this  article  afford  us  another 
method  of  drawing  a  tangent  to  the  ellipse  at  a  given  point. 
Let  P"  be  a  point  at  which  we  wish  to  draw  a  tangent.  Pro- 
duce FT''  to  R,  making  P"Il  =  FP" ;  join  F  and  R.  A  line 
P''T,  drawn  through  P"  _L  to  FE,  will  be  tangent  to  the  ellipse 
at  P". 

89.  To  find  the  condition  thai  the  straight  line  y  ^=-  sx  -\-  c 
must  fulfil  in  order  that  it  may  touch  the  ellipse 

9  9 

+    ^    =    1. 

I  7  0 


/2 


H" 


If  we  consider  the  line  as  a  secant  and  combine  the  equations 
y  =sx  Ar  c, 

a^        0- 

we  obtain  the  co-ordinates  of  the  points  of  intersection. 
Eliminating  y  from  these  equations,  we  have 


—  sa!^c  4-  ab  -y/s^a'^  4-  V^  —  c^  .^  s 

^  = ^ — ^ T^ •  •  •   (1) 

for  the  abscissas  of  the  points  of  intersection.  Now,  when  the 
secant  line  becomes  a  tangent,  these  abscissas  become  equal. 
Looking  at  (1)  we  see  that  the  condition  for  equality  of  ab- 


124  PLANE  ANALYTIC  GEOMETRY. 

scissas  is  that  the  radical  in  the  numerator  shall  disappear ; 
hence 

s^d^  +  ^'^  —  c^  =  0, 
or  s-a-  +b'  =  c'  .  .  .   {2) 

is  the  required  condition. 

Cor.  If  we  substitute  the  value  of  c  drawn  from  (2)  in  the 
equation  of  the  line,  we  have 

y  =  sx  ^  -Vshr  -\-  P    .  .  .   (3) 

for  the  equation  of  the   tangent  to  the  ellijose  in  terms  of  its 
slope. 

90.  To  find  the  locus  generated  by  the  intersection  of  a  tan- 
gent to  the  ellipse  and  a  perpendicidar  to  it  from  a  focus  as 
the  point  of  tangency  moves  around  the  curve. 

The  equation  of  a  straight  line  through  the  focus  (ae,  o)  is 

y  =  s'  (x  —  ae). 
In  order  that  this  line  shall  be  perpendicular  to  the  tangent 
y=.sxJ^  -Vs^a^  +  b^  .  .  .   (1), 
its  equation  must  be 

y=--(x-ae)  .  .  .   (2) 
s 

If  we  now  combine  (1)  and  (2)  so  as  to  eliminate  the  slope 
(s),  the  resulting  equation  will  express  the  relationship  be- 
tween the  co-ordinates  of  the  point  of  intersection  of  these 
lines  in  every  position  they  may  assume ;  hence  it  will  be  the 
equation  of  the  required  locus. 

Transposing  sx  to  the  first  member  in  (l).  and  clearing  (2) 
of  fractions  and  transposing,  we  have 

y  —  sx  =  ^  -yy'shi'^  +  ^^• 
sy  -\-  x  ^  ae. 

Squaring  these  equations  and  adding,  remembering  that 
a^  —  b^  =  ah"^,  Art.  76,  we  have, 

(1  +  s')  (x^-  +  y')  =  (1  +  s')  a', 
or  x~  -\-  y^  =  a^ ;  .  .  .   (3)  ; 


THE  ELLIPSE.  125 

hence,  tJie  circle  constructed  on  the  transverse  axis  of  the  ellipse 
is  the  locus  of  the  intersection  of  the  tangents  cmd  the  perpen- 
dicidars  let  fall  from  the  focus  on  them. 

This  circle  is  known  as  the  Major-Director  circle  of  the 
ellipse.     (See  Fig.  45. ) 

91.  To  find  the  locus  generated  by  the  intersection  of  two 
tangents  which  are  perpendicxdar  to  each  other  as  the  points  of 
tangency  move  around  the  curve. 

The  equation  of  a  tangent  to  the  ellipse  is 


y  ^  sx  -\-  -y/s'^a'^  -{-  b^  .   .   .   (1) 
The  equation  of  a  tangent  perpendicular  to  (1)  is 

2/=-la-+v/4  +  ^';  ...  (2) 
s  \    S' 

hence,  by  a  course  of  reasoning  analogous  to  that  of  the  pre- 
ceding article,  we  have 

a;2  +  ^y2  _  ^2  ^  J2    _     _     _     ^3) 

The  required  locus  is,  therefore,  a  circle  co?icent7'ic  with  the 
ellipse  and  having  its  radius  equal  to  -\/a'^  -{-  b^. 

92.  Two  tarigents  are  drawn  to  the  ellipse  from  a  point  with- 
out ;  required  the  equation  of  the  line  joining  the  points  of 
tangency. 

Let  P'  {x',  ?/),  Fig.  42,  be  the  given  point,  and  let  P''  {x",  y"), 
P2  {x^,  ^2)  be  the  points  of  tangency.  Since  P'  (x',  y')  is  a 
point  common  to  both  tangents,  its  co-ordinates  must  satisfy 
their  equations ;  hence, 

x'x"    I   yV  _  ^ 
a^     ~^    b-" 

a^  b-^ 

Hence  (x",  y")  and  (x2, 2/2)  will  satisfy  the  equation 


^  +  ^/  =  1  .   .  .   (1) 
cr         b- 


126 


PLANE  ANALYTIC   GEOMETRY. 


As  (1)  is  the  equation  of  a  straight  line,  and  is  satisfied  for 
the  co-ordinates  of  both  points  of  tangency,  it  must  be  the 
equation  of  the  straight  line  which  joins  them. 

93.  To  find  the  equation  of  the  polar  of  the  pole  {x\  y'),  with 
regard  to  the  ellipse 


Fig.  42. 


By  the  aid  of  Fig.  42,  and  a  course  of  reasoning  similar  to 
that  of  Art.  49,  the  equation  of  PiP",  the  polar  to  P',  may  be. 
shown  to  be 

x'x    .   y'y  ^  ^ 

CoR.  If  the  polar  of  the  point  P'  (x',  y')  passes  through 
Pi  {^1,  y\),  then  the  polar  of  Pi  {x^,  y^)  will  pass  through 
P'  (x',  y').     (See  Art.  50.) 

94.  To  deduce  the  equation  of  the  ellipse  when  referred  to  a 
pair  of  conjugate  diameters  as  axes. 

A  pair  of  conjugate  diameters  of  the  ellipse  are  those  diam- 


THE  ELLIPSE.  127 

eters  to  which  if  the  ellipse  be  referred  its  equation  will  co7itain 
only  the  second  powers  of  the  variables. 

The  equation  of  the  ellipse  when  referred  to  its  centre  and 
axes  is 

a,  0- 

If  we  refer  the  ellipse  to  a  pair  of  oblique  axes  having  the 
origin  at  the  centre,  we  have,  Art.  33,  Cor.  1, 

X  =  x'  cos  6  -\-  >/  cos  q) 
y  =  x'  sin  0  -\-  y'  sin  cp 

for  the  equations  of  transformation.  Substituting  in  (1),  we 
have 

(a^  sin^  6  -\-b'^  cos^  6)  x"^  +  (^^  sin^  <P  +  ^^  cos^  qp)  y'  ^ 
+  2  (o?-  sin  Q  sin  gp  +  ^^  cos  Q  cos  go)  x'y'  =  a%'^  ...   (2) 

for  the  equation  of  the  ellipse  referred  to  oblique  axes.  But, 
by  definition,  the  equation  of  the  ellipse  when  referred  to  a 
pair  of  conjugate  diameters  contains  only  the  second  powers 
of  the  variables  ;  hence 

a~  sin  6  sin  q>  -\- b'  cos  9  cos  qp  =  0  .  .  .   (3) 

is  the  condition  that  a  pair  of  axes  must  fulfil  in  order  to  be 
conjugate  diameters  of  the  ellipse. 

Making  the  co-efficient  of  x'y'  equal  to  zero  in  (2),  we  have 
after  dropping  accents 

(o-^  sin^  6  -\-b'^  cos^  6)  x'^  -\-  (cr  sin^  <jp  +  ^^  cos^  (p)  y"^  =  a'^b'^ ...  (4) 

for  the  equation  of  the  ellipse  when  referred  to  a  pair  of  con- 
jugate diameters.  This  equation,  however,  takes  a  simpler 
form  when  we  introduce  the  semi-conjugate  diameters.  Mak- 
ing y  =  0  and  x  =  0,  successively,  in  (4),  we  have 


^  a  -    \ 

a-  sin'-^  0  -\-  b'  cos^  0 


a-  sin-  ()p  +  t»^  cos-^  go  } 


^    .  .  .   (5) 


128  PLANE  ANALYTIC  GEOMETRY. 

in  which  a'  and  h'  represent  the  semi-conjugate  axes.     From 
(5),  we  have 

a"  sin^  e  +  1)"  cos2  0  =  ^  ; 

a  - 

a-  sm-  go  +  ^^  cos-  (f  =  -— rr-  . 

Substituting  these  values  of  the  co-efficients  in  (4),  we  have, 
after  reduction, 

—  +  i^  =  1  .  .  .   (6) 

for  the  required  equation. 

Cob.  As  equation  (6)  contains  only  the  second  powers  of 
the  variables,  it  follows  that  each  of  the  two  diameters  to 
which  the  curve  is  referred  will  bisect  all  chords  drawn 
parallel  to  the  other. 

ScHOL.  The  equation  of  condition  for  conjugate  diameters 
(3)  may  be  put  under  the  forms 

tan  6  tan  go  ^ .  .  .  (J) 

a" 

Comparing  this  expression  with  (3)  Art.  81,  we  see  that  the 
same  result  was  obtained  for  the  supplementary  chords  of  an 
ellipse;  hence.  Fig.  40,  if  A'E,',  R'A  be  a  pair  of  supplement- 
ary chords,  then  RP'',  PR",  drawn  through  the  centre  parallel 
to  these  chords,  will  be  a  pair  of  conjugate  diameters.  Again : 
comparing  (7)  with  (1)  Art.  84,  we  see  that  the  same  relation- 
ship was  obtained  for  a  diameter  and  the  tangent  drawn  at 
its  extremity ;  hence.  Fig.  40,  if  P''E  be  a  diameter  and  P"T 
be  a  tangent  drawn  at  its  extremity,  then  PR",  drawn  through 
the  centre  parallel  to  P"T,  is  the  conjugate  diameter  to  RP". 

The  equation  of  condition  (7)  being  a  single  equation  con- 
taining two  unknown  quantities  (tan  9,  tan  (f),  we  may 
assume  any  value  we  please  for  one  of  them,  and  the  equation 
will  make  known  the  value  of  the  other ;  hence,  in  the  ellipse 
there  are  an  infinite  number  of  pairs  of  conjugate  diameters. 


THE  ELLIPSE. 


129 


95.    To  find  the  equation  of  a  conjugate  diameter. 
Let  P"R,  RT'  be  a  pair  of  conjugate  diameters.     We  wish 
to  find  the  equation  of  RT'. 

Y 


Fig.  43. 


The    equation   of   the    tangent   line   P"T,   drawn   through 
V"  (x",  y")  is 


a^ 


+ 


yy 


1. 


By  Art.  94,  SchoL,  the  diameter  P'R'  is  parallel  to  P"T ; 
hence  its  equation  must  be  the  same  as  that  of  the  tangent, 
the  constant  term  being  zero. 


y  = 


XX 

1   yy 

a'' 

'    h^ 



h'^x" 
r;^ 

=  0 


(1) 


ahf 


(2) 


is  the  equation  of  a  diameter  expressed  in  terms  of  the  co- 
ordinates of  the  extreinity  of  its  conjugate  diameter. 

CoR.    Let  s  represent  the  slope  of  the  diameter  P''R  ;  then, 
from  (2) 

crij  a^    s 


130  PLANE  ANALYTIC  GEOMETRY. 


OD 
DP" 

x"       1 

y"     s 

hence  we  have 

y  = 

X    . 

(3) 

for  the  equation  of   a  diameter  in  terms  of  the  slope  of  its 
conjugate  diameter. 

96.  To  find  the  co-ordinates  of  either  extremity  of  a  diam- 
eter, the  co-ordinates  of  one  extremity  of  its  conjugate  diameter 
being  given. 

Let  P"E  and  R'P',  Fig.  43,  be  a  pair  of  conjugate  diameters. 
Let  {x",  y")  be  the  co-ordinates  of  P".  We  wish  to  find  the 
co-ordinates  (x',  y')  of  P'  in  terms  of  the  co-ordinates  of  P." 

The  equation  of  condition  that  P'  {x',  y')  shall  be  on  the 
diameter  FR'  is,  Art.  95,  (1) 


a''      '      b'^ 
Since  P'  (x\  y')  is  on  the  ellipse,  we  have  also 

a""        b^ 

Eliminating  y'  and  x',  successively,  from  these  equations, 
we  find 

x'  =^-  if  and  /  =  ±  -  oc!'. 
b  a 

These  expressions,  taken  with  the  upper  signs,  are  the  co- 
ordinates of  P' ;  taken  with  the  lower  signs,  they  are  the 
co-ordinates  of  E.'. 

97.  To  shoiv  that  the  sum  of  the  squares  on  any  pair  of 
semi-conjugate  diameters  is  equivalent  to  the  sum  of  the  squares 
on  the  semi-axes. 

Let  V"  (x",  y")  and  P'  (x',  /),  Fig.  43,  be  the  extremities 


THE  ELLIPSE.  131 

of  any  two  semi-conjugate  diameters.    Let  OP"  =  a',  OP'  =  b' ; 
then,  from  the  triangles  ODP",  ODT',  we  have, 

a"'  =  x'"'^y"-'  ...   (1) 
and  b'^  =  x'^  +  y"'  •  •  •  (2) 


But,  Art.  96,       x'^  =  -^  ij"% 


and  y'^  =  —  x'"^ ; 


hence  b'^  =  —  t/'^  +  —  x'""  .  .  .  (3). 

Adding  (1)  and  (3),  we  have 

\  a^         b^  J 

but  ^  +  2^  =  1 . 

a"  ^  b^ 

hence,  a'^  _|_  ^'2  ^  ^2  _|_  ^2  _  _  _  (^4^) 


98.  To  show  that  the  parallelogram  constructed  on  any  tivo 
conjugate  diameters  is  equivalent  to  the  rectangle  constructed 
on  the  axes. 

Let  P"R  (  =  2a'),  FR'  (  =  2b'),  Pig.  44,  be  any  two  conju- 
gate diameters.     To  prove  that  area  CTC'T'  =  area  BB'H'H. 

The  area  of  the  parallelogram  OP"TR'  is 

OP'  X  P"P- 

From  the  figure  P"P  =  OP"  sin  P"OP/ 

=  a'  sin  (180°  -  {(p  —  6)  )  ^  a'  sin  ((p  -  0)  ; 
.:  area  of  OP"TP/  =  a'b'  sin  (cp  -  $)  .  .  .  (1) 


132' 


PLANE  ANALYTIC  GEOMETRY. 


From  the  triangles  OD'P',  ODP'',  we  have 

D'P'       y'         hx"        .     a 
—  -^^  — •  sm  Q 


sm  (jD  = 


OD' 


OP' 


y 


ah' 


COS  QD  :=  = = 

OP'  b' 


ail  /,        X 

— ^  :  cos  Q  = — - 

hV    '  a' 


Hence 


sin  (qp  —  ^)  =  sin  go  cos  0  —  cos  (jd  sin  B 


ay- 


hx" 

aa'h'   '    la'h' 

aa'b'b 

aW 
aba'b' 

ab 


THE  ELLIPSE. 


133 


Substituting  this  value  in  (1)  and  multiplying  through  by 

4,  we  have 

area  OP"TR'  X  4  =  4  a^> ; 
area  CTC'T'  =  area  BB'H'H. 


I.e., 


99,  To  shoiv  that  the  ordinate  of  any  point  on  the  ellipse  is 
to  the  ordinate  of  the  corresponding  point  on  the  circumscribing 
circle  as  the  semi-conjugate  axis  of  the  ellipse  is  to  the  semi- 
transverse  axis. 


R 

/-^ 

B 

A 

?" 

«M 

/ 

Xv 

X 

D 

X 

[ 

D' 

J 

Jj 

B' 
Fig.  45. 

Let  DP',  DP''  be  the  ordinates  of  the  corresponding  points 

P'  (x',  y')  and  P"  (x",  y"). 

Since  P'  (x',  y')  is  on  the  ellipse,  we  have 

y-  =  —  (a^  —  X-). 
a- 

Since  Y"  (x",  y")  is  on  the  circle  whose  radius  is  a,  we  have 

y"^-  =  a2  _  a;"l 

Dividing  these  equations,  member  by  member,  we  have 

fO  7  9 

^  =  —  ,     (since  X  =x    )  ; 
y  -^       a- 

.:  y' :  y"  ::  b  :  a. 


134 


PLANE  ANALYTIC  GEOMETRY. 


Similarly  we  may  prove  that 

where  Xi  is  the  abscissa  of  any  point  on  the  ellipse,  and  x^  is 
the  corresponding  abscissa  of  a  point  on  the  inscribed  circle. 

100.  The  principles  of  the  preceding  article  give  us  a 
method  of  describing  the  ellipse  by  points  when  the  axes  are 
given. 

From  0,.  Fig.  45,  as  a  centre  with  radii  equal  to  the  semi- 
axes  OA,  OB  describe  the  circles  A'RA,  BOB'.  Draw  any 
radius  OR  of  the  larger  circle,  cutting  the  smaller  circle  in  M  ; 
draw  MIST  ||  to  OA',  cutting  the  ordinate  let  fall  from  E  in  jST  ; 
N  is  a  point  of  the  ellipse.  Since  MN  is  parallel  to  the  base 
of  the  triangle  ED'O,  we  have 

D'N  :  D'R  ::  OM  :  OR ; 
i.  e.,  y'  -.y"  ■.-.h-.a; 

hence,  the  construction. 

101.  To  show  that  the  area  of  the  ellipse  is  to  the  area  of 

the  circumscribing  circle  as  the  semi-minor  axis  of  the  ellijjse  is 

to  its  semi-major  axis. 

Pa 

Pi 


Fig.  46. 


THE  ELLIPSE.  135 

Inscribe  in  the  ellipse  any  polygon  ARRiE,2E'3l^4A',  and 
from  its  vertices  draw  the  ordinates  RD,RiDi,  etc.,  producing 
them  upward  to  meet  the  circle  in  P,  P^,  Pg,  etc.  Joining 
these  points  we  form  the  inscribed  polygon  APP1P2P3P4A'  in 
the  circle. 

Let  (x,  y^),  (x',  y{),  {x",  y<^  etc.,  be  the  co-ordinates  of 
P,  Pi,  Pg,  etc.,  and  let  (a;,  ?/),  (x' ,  y'),  (x",  y"),  etc.,  be  the  co-ordi- 
nates of  the  corresponding  points  R,  E-i,  Eg?  etc.,  of  the  ellipse. 

Then  Area  RDDiRj  =  (x  -  x')  ^  ^^^ 

Area  PDDiPi  =  {x  -  x')  l^+ll . 

hence  Area  RDDiRi  _   y +/ 

Area  PBD^Pi       2/0  +  2/i ' 

But,  Art.  99,  ^  =  -  and  i^  ==  - ; 
yo       a  yi       a 

2/0+2/1       »  ' 

Hence  Area  RDDiRi  _  ^ 

Area  PDDjPi  ~  a  ' 

We  may  prove  in  like  manner  that  every  corresponding  pair 
of  trapezoids  bear  to  each  other  this  constant  ratio ;  hence, 
by  the  Theory  of  Proportion,  the  sum  of  all  the  trapezoids  in 
the  ellipse  will  bear  to  the  sum  of  all  the  trapezoids  in  the 
circle  the  same  ratio.  Representing  these  sums  by  'M  and 
ST,  respectively,  we  have 

^t        b 


ST        a 

As  this  relationship  holds  true  for  any  number  of  trape- 
zoids, it  holds  true  for  the  limits  to  which  the  sum  of  the 
trapezoids  of  the  ellipse  and  the  sum  of  the  trapezoids  of 
the  circle  approach   as   the    number   of  trapezoids    increase. 


136  PLANE  ANALYTIC  GEOMETRY. 

But  these  limits  are  the  area  of  the  ellipse  and  the  area  of 
the  circle ;  hence 

area  of  ellipse   h 

area  of  circle         a 

Cor.    Since  the  area  of  the  circle  is  tt  a^,  we  have 

area  of  ellipse  h 

TT  a^  a 

.*.  area  of  ellipse  =  tt  ab. 
Since  -k  a^ : -k  ah  :  :  ir  ah  :  tt  V^ , 

we  see  that  the  area  of  the  ellipse  is  a  mean  proportional  be- 
tween the  areas  of  the  circumscribed  and  inscribed  circles. 

EXAMPLES. 

1.  What  must  be  the  value  of  c  in  order  that  the  line 
y  =  2  X  -\-  c  may  touch  the  ellipse 

Ans.     G  =  5. 

2.  The  semi-transverse  of  an  ellipse  is  10 ;  what  must  be 
the  value  of  the  semi-conjugate  axis  in  order  that  the  ellipse 
may  touch  the  line  2?/  +  a;  —  14  =  0? 

Ans.     b  =  V24. 

3.  What  are  the  equations  of  the  tangents  to  the  ellipse 

^''  _i_  f"  —  1 

whose  inclination  to  X-axis  =  45°  ? 

4.  The  locus  of    the  intersection  of  the  tangents   to  the 

2  2 

ellipse  -—  +  ^  =  1 

a-         b'^ 

drawn  at  the  extremities  of  conjugate  diameters  is  an  ellipse  ; 

required  its  equation. 

^^^-     -^  +  li  =  2. 
a^       b^ 


THE  ELLIPSE.  137 

5.  Tangents  are  drawn  from  the  point  (0,  8)  to  tlie  ellipse 

^  +  y^  =  1 ; 

required   the    equation   of    the   line    joining  the    points    of 
tangency.  Ans.     8  ?/  —  1  =  0. 

Eequired  the  polar  of  the  point  (5,  6)  with  respect  to  the 
following  ellipses : 

6.  x^  +  3,f  =  <d.  7.   1  +  ^=1. 

8.    ^  -L  l!  =  1. 

9.  What  are  the  polars  of  tlie  foci  ? 

A71S.     a;  =  -j^  — 
.  e  • 

10.  What  is  the  pole  of  ?/  =  3  x  +  1  with  respect  to 

_^ [_  J/_  ^  ]^? 

4   "^    9  ' 

Ans.     (-  12,  9). 

11.  The  line  3  ?/  =  5  x  is  a  diameter  of 

^^    9   ~     ' 
required  the  equation  of  the  conjugate  diameter. 

Ans.     20  y  +  27  a;  =  0. 

12.  A  pair  of  conjugate  diameters  in  the  ellipse 

^-u  i^  =  1 
16         9 

3  3 

make  angles  whose  tangents   are    -j  and  —    j,  respectively, 

with  the  X-axis  ;  required  their  lengths. 

13.  What  is  the  area  of  the  ellipse 


5_  4.  ^  =  19 
4   ^  10 


Ans.     2  TT  VlO. 


138  PLANE  ANALYTIC   GEOMETRY. 

14.  The  minor  axis  of  an  ellipse  is  10,  and  its  area  is  equal 
to  the  area  of  a  circle  whose  diameter  is  16 ;  what  is  the  length 
of  the  major  axis  ?  Ans.     25 J. 

15.  The  minor  axis  of  an  ellipse  is  6,  and  the  sum  of  the 
focal  radii  to  a  point  on  the  curve  is  16  ;  required  the  major 
axis,  the  distance  between  the  foci,  and  the  area. 

GENERAL    EXAMPLES. 

1.  What  is  the  equation  of  the  ellipse  which  passes  through 
(2,  4)  ( —  2,  4),  the  centre  being  at  the  origin  ? 

2.  The  major  axis  of  an  ellipse  is  =  18,  and  the  point 
(6,  4)  is  on  the  curve ;  required  the  equation  of  the  ellipse. 

1  13 

3.  The  lines  2/  =  —  9  ^  +  6  and  y  =  q^x  -\-  -^  are  supplemen- 
tal chords  drawn  from  the  extremities  of  the  transverse  axis 
of  an  ellipse  ;  required  the  equation  of  the  ellipse. 

4.  The  minor  axis  of  an  ellipse  is  =  12,  and  the  foci  and 
centre  divide  the  major  axis  into  four  equal  parts ;  required 
the  equation  of  the  ellipse. 

5.  Assuming  the  equation  of  the  ellipse  show  that  the 
sum  of  the  distances  of  any  point  on  the  ellipse  from  the  foci 
is  constant  and  =  to  the  transverse  axis. 

6.  The  sub-tangent  for  a  point  whose  abscissa  is  2  is  =  6 

in  an  ellipse  whose  eccentricity  is  - ;    required  the  equation 

of  the  ellipse.  .  a:^        t/^  ^ 

16  "^  15  ~ 

7.  What  are  the  equations  of  the  tangents  to 

9        25 
which  form  with  the  X-axis  an  equilateral  triangle  ? 

8.  Show  that  the  tangents  drawn  at  the  extremities  of  any 
chord  intersect  on  the  diameter  which  bisects  that  chord. 


THE  ELLIPSE.  139 

9.  What  are  the  equations  of  the  tangents  drawn  at  the 
extremities  of  the  latus-rectum  ? 

10.  Show  that  the  pair  of  diameters  drawn  parallel  to  the 
chords  joining  the  extremities  of  the  axes  are  equal  and 
conjugate. 

11.  A  chord  of  the  ellipse 

^-u^  =1 
16         9 

passes  through  the  point   (2,  3)  and  is  bisected  by  the  line 
y  —  a;  =  0  ;  required  the  equation  of  the  chord. 

12.  What  are  the  equations  of  the  pair  of  conjugate  diam- 
eters of  the  ellipse  16y^  -{-9x^  =  144  which  are  equal  ? 

13.  Show  that  either  focus  of  an  ellipse  divides  the  major 
axis  in  two  segments  whose  rectangle  is  equal  (a)  to  the 
rectangle  of  the  semi-major  axis  and  semi-parameter ;  (b)  to 
the  square  of  the  semi-minor  axis. 

14.  Show  that  the  rectangle  of  the  perpendiculars  let  fall 
from  the  foci  on  a  tangent  is  constant  and  equal  to  the  square 
of  the  semi-minor  axis. 

15.  A  system  of  parallel  chords  which  make  an  angle  whose 
tangent  =  2  with  the  X-axis  are  bisected  by  the  diameter  of 
an  ellipse  whose  semi-axes  are  4  and  3 ;  required  the  equation 
of  the  diameter. 

16.  Show  that  the  polar  of  a  point  on  any  diameter  is 
parallel  to  the  conjugate  diameter. 

17.  Find  the  locus  of  the  vertex  of  a  triangle  having  given 
the  base  =  2  a,  and  the  product  of  the  tangent  of  the  angles 

at  the  base      =  — - . 

Ans.     Irx^  -\-  ^if"  =  Wa^. 


140  PLANE  ANALYTIC  GEOMETRY. 

18.  Find  the  locus  of  the  vertex  of  a  triangle  having  given 
the  base  =2  a,  and  the  sum  of  the  sides  =2  6. 

Ans.     —  -\ ^ —  =  1. 

19.  Find  the  locus  of  the  intersection  of  the  ordinate  of 
the  ellipse  produced  with  the  perpendicular  let  fall  from  the 
centre  on  the  tangent  drawn  at  the  point  in  which  the  ordi- 
nate cuts  the  ellipse. 

20.  Find  the  locus  generated  by  the  intersection  of  two 
tangents  drawn  at  the  extremities  of  two  radii  vectores  (drawn 
from  centre)  which  are  perpendicular  to  each  other. 

Ans.     a'^y^  -\-  b^x^  =  a^  h'^  -\-  h^a*". 

21.  A  line  of  fixed  length  so  moves  that  its  extremities 
remain  in  the  co-ordinate  axes  ;  required  the  locus  generated 
by  any  point  of  the  line. 

22.  The  angle  AOP"  =  q>  (Fig.  45)  is  called  the  eccentric 
angle  of  the  point  P'  {x\  y')  on  the  ellipse.  Show  that  (x',  y') 
=  (ct  cos  q>,  h  sin  gp)  and  from  these  values  of  the  co-ordinates 
deduce  the  equation  of  the  ellipse. 

23.  Express  the  equation  of  the  tangent  at  (x",  y")  in  terms 
of  the  eccentric  angle  of  the  point. 

Ans.     -  cos  (p  -\-  —  sin  (p  =  1. 
a  b 

24.  If  (x',  y'),  (x",  y")  are  the  ends  of  a  pair  of  conjugate 
diameters  whose   eccentric   angles   are   (p   and   tp',  show  that 

q>'  -q>  =  90°. 


THE  HYPERBOLA. 


141 


CHAPTER  VIII. 
THE    HYPERBOLA. 

102.  The  hyperbola  is  the  locus  of  a  point  so  moving  in  a 
plane  that  the  difference  of  its  distances  from  two  fixed  points 
is  always  constant  and  equal  to  a  given  line.  The  fixed 
points  are  called  the  Foci  of  the  hyperbola.  If  the  points 
are  on  the  given  line  produced  and  equidistant  from  its 
extremities,  then  the  given  line  is  called  the  Transverse  Axis 
of  the  hyperbola. 

103.  To  deduce  the  equation  of  the  hyperbola,  given  the  foci 
and  the  transverse  axis. 


Fig.  47. 


Let  F,  F'  be  the  foci,  and  AA'  the  transverse  axis.     DraAv 
OYlto  A  A' at  its  middle  point,  and  take   OY,  OX  as  the 


142  PLANE  ANALYTIC  GEOMETRY. 

co-ordinate  axes.  Let  P  be  any  point  of  the  curve.  Draw 
PF,  PF' ;  draw  also  PD  ||  to  OY. 

Then  (OD,  DP)  =  {x,  y)  are  the  co-ordinates  of  P. 

Let  AA'  =2  a,  FF'  =  2  OF  =  2  0F'=  2  c,  FP  =  r  and  FT 
=  /. 

From  the  right  angled  triangles  FPD  and  F'PD,  we  have 


r  =  Vy^  +  (^  —  <^y  and  /  =  ■yjy'^  -\-  (x  -\-  cy  .  .  .   (a) 
From  the  mode  of  generating  the  curve,  we  have 

r'  -r  =  2a. 
Hence,  substituting, 


^f  +  (^  +  cf  -^y^J^(x-cy  =  2a',  .  .  .   (1) 
or,  clearing  of  radicals  and  reducing,  we  have 

(c"  —  oF)  x^  —  a^y'^  =  a?  {c~  —  or)  .  .  .   (2) 

for  the  required  equation.     This   equation,  like  that  of  the 
ellipse  (see  Art.  75),  may  be  put  in  a  simpler  form. 
Let  c"  -a""  =h^  .  .  .  (3) 

This  value  in  (2)  gives,  after  changing  signs, 
ahf-  -  Vx'  =  -  o?h-,  ...   (4) 
or,  symmetrically, 

^-f^  =  l.  •  •   (5) 
a-        b- 

for  the  equation  of  the  hyperbola  when  referred  to  its  centre 
and  axes. 

Let  the  student  discuss  this  equation.  (See  Art.  14) 
Cor.  1.       li  h  =  a  in  (5),  we  have 

x^  -,f  =  a?  ,  .  .   (6) 

The  curve  represented  by  this  equation  is  called  the  Equi- 
lateral Hyperbola.  Comparing  equation  (6)  with  the  equation 
of  the  circle 

a?^  +  2/^  =  «^ 
%ve  see  that  the  equilateral  hyperbola  bears  the  same  relation  to 
the  comvion  hyperbola  that  the  circle  bears  to  the  ellipse. 


THE  HYPERBOLA.  143 

Cor.  2.    If  (x',  y')  and  (x",  y")  are  the  co-ordinates  of  two 
points  on  the  curve,  we  have  from  (4) 

y'^  =  il  (x'2  -  a")  and  y"^  =  -^  (cc"^  _  a^)  • 
Oj  cc 

hence      y'^  :  y"" ::  (x'  —  a)  (x  -{-  a)  :  {x"  —  a)  (x"  +  «)  ; 

i.e.,  the  squares  of  the  ordinates  of  any  two  points  on  the 
hyperbola  are  to  each  other  as  the  rectangles  of  the  segments  in 
tvhich  they  divide  the  ti-ansverse  axis. 

Cob.  3.    By  making  x  =  x'  —  a  and  y  =■]/  va.   (4)  we  have 
after  reducing  and  dropping  accents, 

aY  -V'x^-\-2  aWx  =  0  ...   (7) 

for  the  equation  of  the  hyperbola,  A'  being  taken  as  origin. 

104.    From  equation  (3)  Art.  103,  we  have 


^»  =  J_  Vc2  -  a'. 

Laying  this  distance  off  above  and  below  the  origin  on  the 

Y-axis,  we  have  the  points  B,  B',  Fig.  47,  Art.  103.      The  line 

BB'  is  called  the   Co^mugate  Axis  of  the   hyperbola.     The 

jDoints  A  and  A'  are  called  the  Vertices  of  the  curve.     The 

point  0  bisects  all  lines  drawn  through  it  and  terminating  in 

the  curve  ;    for  this  reason  it  is  called  the  Centre  of  the 

hyperbola. 

The  ratio         -^Ti^jITfl        r 

^''  +^    =~  =  e.     See  (3)  Art.  103  ..  .   (1) 
a  a 

is  called  the  Eccentricity  of  the  hyperbola.  This  ratio  is 
evidently  >  1.  The  value  of  c  =  -|-  Va^  +  b^  measures  the 
distance  of  the  foci  F,  F'  from  the  centre. 

If  b  =  a  in  (1),  we  have  e  =  V2  for  the  eccentricity  of  the 
equilateral  hyperbola. 

105.    To  find  the  values  of  the  focal  radii,  r,  r'  of  a  point 
on  the  hyperbola  in  terms  of  the  abscissa  of  the  point. 
From  equations  (a)  Art.  103,  we  have 

?'  =  V  2/^  +  (^  —  (^Y ' 


144 


PLANE  ANALYTIC  GEOMETRY. 


From  the  equation  of  the  hjqoerbola,  (4)  Art.  103,  we  have 

y^  =  — -  (x^  —  a^)  =  —  x^  —  b^. 
Hence,  substituting 


=v/: 


p 


x"^  —  b^  -\-  x^  —  2  ex  -\-  c% 


V        a- 


x^  —  2  ex  -}-  c^  —  b^, 


=  d^x^  -2cx  +  a'-,    Art.  104  (1), 


=  —X  —  a\ 
a 

hence  r  =  ex  —  a  .  .  .  (V) 

Similarly,  we  find 

r'  =  ex  -{-  a  .  .  .   (2) 

106.    To  construct  the  hyperbola  having  given  the  transverse 
axis  and  the  foci  of  the  curve. 


Fig.  48. 


First  Method.  —  Let  A  A'  be  the  transverse  axis  and  F,  F',  the 
foci.    Take  a  straisfht-edge  ruler  whose  length  is  L  and  attach 


THE  HYPERBOLA.  145 

one  of  its  ends  at  F'  so  that  the  ruler  can  freely  revolve  about 
that  point.  Cut  a  piece  of  cord  so  that  its  length  shall  be 
=  L  —  2  a,  and  attach  one  end  to  the  free  end  of  the  ruler, 
and  the  other  end  to  the  focus  F.  Place  the  ruler  in  the 
position  indicated  by  the  full  lines,  Fig.  48,  and  place  the 
point  of  a  pencil  in  the  loop  formed  by  the  cord.  Stretch 
the  cord,  keeping  the  point  of  the  pencil  against  the  edge  of 
the  ruler.  If  we  now  revolve  the  ruler  upward  about  F',  the 
point  of  the  pencil,  kept  firmly  pressed  against  the  ruler, 
will  describe  the  arc  AP'  of  the  hyperbola.  By  fixing  the 
end  of  the  ruler  at  F,  we  may  describe  an  arc  of  the  other 
branch.  It  is  evident  in  this  process  that  the  difference  of 
the  distances  of  the  point  of  the  pencil  from  the  foci  F',F, 
is  always  equal  to  2  a. 

Second  Method.  —  Take  any  point  D  on  the  transverse  axis. 
Measure  the  distances  A'D,  AD.  With  F'  as  a  centre  and  A'D 
as  a  radius  describe  the  arc  of  a  circle  ;  with  F  as  a  centre  and 
AD  as  a  radius  describe  another  arc.  The  intersection  of 
these  arcs  will  determine  two  points,  Pi,  Po,  of  the  curve.  By 
interchanging  centres  and  radii  we  may  locate  the  points  Pi, 
Rs)  on  the  other  branch.  In  this  manner  we  may  determine  as 
many  points  as  the  accuracy  of  the  construction  may  require. 

107.  To  find  the  latus-rectuvi  or  parameter  of  the  hyperbola. 
The  Latus-Pectum,  or  Parameter  of  the  hyperbola,  is  the 
double  ordinate  passing  through  either  focus. 

Making  x^=  ^^  Vft^  -|-  b'^  in  the  equation  of  the  hyperbola 

IP- 

2/"  =  ^r  (^^  —  «^)» 
a- 

ip-  2  b"^ 

we  have  y  =  —  .-.  2  y 


a  a 

Forming  a  proportion  from  this  equation,  we  have 
2y:2b::b:a; 
.:2y:2h::2b:2a; 

i.e,  the  latus-rectum  of  the  hyperbola  is  a  third  pjroportional  to 
the  axes. 


146 


PLANE  ANALYTIC  GEOMETRY. 


108.  The  equation  of  the  ellipse  when  referred  to  its  centre 
and  axes  is 

a'^y'^  +  ^^^^  =  f'^^"- 

The  equation  of  the  hyperbola  when  referred  to  its  centre 

and  axes  is 

ay  —  b-x"^  =  —  a~h'^. 

Comparing  these  equations,  we  see  that  the  only  difference 
is  in  the  sign  of  b"^.  If,  therefore,  in  the  various  analytical 
expressions  we  have  deduced  for  the  ellipse,  we  substitute 
—  b'^  for  b^,  or,  what  is  the  same  thing,  +  &  V—  1  for  b,  we 
will  obtain  the  corresponding  analytical  expressions  for  the 
hyperbola. 

109.  To  deduce  the  equation  of  the  conjugate  hyperbola. 
T\vo  hyperbolas  are  Conjugate  when  the  transverse  and  con- 
jugate axes  of  one  are  respectively  the  conjugate  and  trans- 
verse axes  of  the  other. 


Thus  in  Fig.  49,  if  AA'  be  the  transverse  axis  of  the  hyper- 
bola which  has  BB'  for  its  conjugate  axis,  then  the  hyperbola 
which  has  BB'  for  its  transverse  axis  and  AA'  for  its  conjugate 


THE  HYPERBOLA.  147 

axis  is  its  conjugate;  and,  conversely,  the  hyperbola  whose 
transverse  axis  is  BB'  and  conjugate  axis  is  AA'  has   for  its 
conjugate   the   hyperbola  whose  transverse  axis  is  AA'  and 
whose  conjugate  axis  is  BB'. 
We  have  deduced.  Art.  103,  (5), 

£!  -  l!  =  1  .  .  .  (1) 

for  the  equation  of  the  hyperbola  whose  transverse  axis  lies 
along  the  X-axis.  We  wish  to  find  the  equation  of  its  conju- 
gate. It  is  obvious  from  the  figure  that  the  hyperbola  which 
has  BB'  for  its  transverse  axis  and  AA'  for  its  conjugate  axis 
bears  the  same  relation  to  the  Y-axis  as  the  hyperbola  whose 
transverse  axis  is  AA'  and  conjugate  axis  is  BB'  bears  to  the 
X-axis  ;  hence,  changing  a  to  h  and  b  to  a,  x  to  y  and  y  to  x 
in  (1),  we  have 

li  _  ^  =  1 

or  4  -  ^  =  -  1  •  •  •   (2) 

for  the  equation  of  the  conjiigate  hyperbola  to  the  hyperbola 
whose  equation  is  (1). 

Comparing  (1)  and  (2)  we  see  that  the  equation  of  any 
hyperbola  and  that  of  its  conjugate  differ  only  in  the  sign  of 
the  constant  term. 


CoR.  Since  V^^  +  ^-^  =  V«^  -f-  b^,  the  focal  distances  of 
any  hyperbola  and  those  of  its  conjugate  are  equal. 

The  eccentricities  of  conjugate  hyperbolas,  however,  are 
not  equal.  For  the  hyperbola  whose  semi-transverse  axis  is 
a  and  semi-conjugate  axis  is  h,  we  have 


Art.  104,  (1)  e=^^'  +  ^'. 
a 

For  its  conjugate  hyperbola,  we  have 

,_  Vfr  +  b-" 


148  PLANE  ANALYTIC  GEOMETRY. 

EXAMPLES. 

Find  the  semi-axes,  the  eccentricity  and  the  latus-rectum 
of  each  of  the  following  hyperbolas  : 

1.    9  ^2  _  4  ^2  _  _  og_  b.    3^/  -2x^  =  12. 

6.    ay-  —  hx^  =  —  ab. 

4        ^ 

8.    y'  —  mx'  =  n. 

Write  the  equation  of  the  h3^perbola  having  given : 

9.    The  transverse   axis  =  12 ;    the   distance   between   the 
foci  =  16. 

Ans. ^  =  1 

36       28 


2. 

X-         2/-  _  1 
4          9 

3. 

y'^  —  16  X'  =  —  16. 

4. 

4  cc^  —  16  y-  =  —  64, 

10.    The  transverse  axis  =  10;  parameter  =  8. 


/2 


Ans. i^  ^  1. 

25       20 

11.  Semi-conjugate  axis  =  6  ;  the  focal  distance  =  10. 

Ans. -^  =  1. 

64       36 

12.  The  equation  of  the  conjugate  hyperbolatoa;^— 3  y-  =  6. 

Ans.     cc2  —  3  2/'  +  6  =  0. 

13.  The   conjugate  axis   is  10,  and  the  transverse  axis  is 
double  the  conjugate. 

.            x^         %r       -, 
Ans. -^  =1  1. 

100       25 

14.  The  transverse  axis  is  8,  and  the  conjugate   axis  =  \ 

distance  between  foci. 

Ans.     i^-V  =  i. 
16        16 


THE  HYPERBOLA.  149 

15.    Given  the  hyperbola 

^  _  J^  =  1- 
10         4  ' 

required  the  co-ordinates  of  the  point  whose  abscissa  is  double 

its  ordinate. 


-•  (Vf '  \/f ) 


16.  Write  the  equation  of  the  conjugate  hyperbola  to  each 
of  the  hyperbolas  given  in  the  first  eight  examples  above. 

17.  Given  the  hyperbola  9  ?/^  —  4  a;^  =  —  36 ;  required  the 
focal  radii  of  the  point  whose  ordinate  is  =  1  and  abscissa 
positive. 

18.  Determine  the  points  of  intersection  of 

—  —  -^^—  =  1,  and  -—  +  -^  =  1. 
4  9  '  16        16 

110.  To  deduce  the  iwlar  equation  of  the  hyperbola,  either 
focus  being  taken  as  the  pole. 

Let  us  take  F  as  the  pole,  Fig.  47. 

Let  (FP,  PFD)  =  (?•,  6)  be  the  co-ordinates  of  any  point  P 
on  the  curve.     From  Art.  105,  (1),  we  have 

1^^  =  r  =  ex  —  a  .  .  .   (1) 
From  Fig.  47,    OD  =  OF  +  FD ; 
i.e.,  X  =  ae  -\-  r  cos  6. 

Substituting  this  value  in  (1)  and  reducing,  we  have 

r  =  _   ''  (^  -  ^')    ...   (2) 
1  _  e  cos  ^  ^  ^ 

for  the  polar  equation  of  the  hyperbola,  the  right  hand  focus 
being  taken  as  the  pole. 

Similarly  from  Art.  105,  (2),  we  have 

a(l-  e^) 
1  —  e  cos  9 

for  the  polar  equation,  the  left  hand  focus  being  the  pole. 


r-  =  ^^ ^  ■  •  •   (3) 

1  —  e  cos  9 


150 

PLANE  ANALYTIC  GEOMETRY. 

Cor. 

If  e  =  0,  r  =  -  a  -  ae  =  -  FA', 

r'  =  a  -\-  ae  =  F'A. 

If  ^  = 

90°, 

,        2        a"e^  —  a^        b^  .  ,  ^ 

r  =  —  a  -\-  ae   = =  —  =  semi-Latus  rectum. 

a  a 

r'  ^  a  —  ae^  = = =  semi-latus  rectum. 

a  a 

lie  =  180°,  r  =  -  a  +  ae  =  FA, 

r'  =  a  —  ae=  —  F'A'. 

If  6  =  270°,  ?•  =  —  a  -\-  ae^  =  —  =  semi-latus  rectum. 

a 

b^ 

r'  ^  a  —  ae^  = =  semi-latus  rectum. 

a 

111.  To  deduce   the  equation  of   condition  for  the  stipple- 
mentary  chords  of  the  hyperbola. 

By  a  method  similar  to  that  of  Art.  81,  or  by  placing  —  h'^ 
for  y^  in  (3)  of  that  article,  we  have 

ss'  =  ^;...    (1) 

hence,  the  product  of  the  slopes  of  any  pair  of  supplementary 

chords  of  an  hyperbola  is  the  same  for  every  pair. 

CoR.  If  a.  =  6,  we  have 

ss   =  1,  or,  s  =  — , 
s 

.'.  tan  «  =  cot  «' ; 
hence,  the  sum  of  the  ttvo  acute  angles  ivhich  any  p)air  of  sup- 
pjlementary  chords   of  an  equilateral  hyperbola  make  with  the 
X-axis  is  equal  to  90°. 

112.  T'o  dedaice  the  equation  of  the  tangent  to  the  hyperbola. 
By  a  method  entirely  analogous    to  that   adopted    in   the 

circle,  or  ellipse,  or  parabola,  Arts.  41,  82,  57 ;  or  substituting 
—  ^2  for  7.2  in  (5)  of  Art.  82,  we  find 

'-   yl=  1  ...   (1) 

to  be  the  equation  of  the  tangent  to  the  hyperbola. 


THE  HYPERBOLA.  151 

113.  2^0  deduce  the  value  of  the  sub-tangent. 

By  operating  on  (1)  of  the  preceding  article  (see  Art.  83), 
we  find 

Sub-tangent  =  x" —  = . 

114.  Tlie  slope  of  a  line  passing  through  the  centre  of  an 
hyperbola  (0,  0)  and  the  point  of  tangency  {x",  y")  is 

t  =  y:L. 

x" 
The  slope  of  the  tangent  is,  Art.  112,  (1) 

t'  =  ^    ^ 
~   a^  '  y"' 

Multiplying  these  equations,  member  by  member,  we  have 

ttf^^.  .  .  (1) 
a^ 

Comparing  (1)  of  this  article  with  (1)  of  Art.  Ill,  we  find 

ss'  =  tt'  .  .  .  (2) 

Hence,  the  line  from  the  centre  of  the  hyperbola  to  the 
point  of  tangency  and  the  tangent  enjoy  the  property  of  being 
the  supplemental  chords  of   an   hyperbola  whose   semi-axes 

bear  to  each  other  the  ratio  -  • 

a 

CoR.  If  s  ^  t,  then  /  =  t' ;  i.e.,  if  one  supplementary 
chord  of  an  hyperbola  is  parallel  to  a  line  drawn  through  the 
centre,  then  the  other  supplementary  chord  is  parallel  to  the 
tangent  drawn  to  the  curve  at  the  point  in  which  the  line 
through  the  centre  cuts  the  curve. 

115.  The  preceding  principle  affords  us  a  simple  method 
of  drawing  a  tangent  to  the  hyperbola  at  any  given  point  of 
the  curve. 


152 


PLANE  ANALYTIC   GEOMETRY. 


Fig.  50. 


Let  P'  be  auy  point  at  which  we  wish  to  draw  a  tangent- 
Join  P'  and  0,  and  from  A'  draw  A'C  ||  to  P'O ;  join  C  and  A. 
The  line  P'T,  drawn  from  P'  ||  to  CA  will  be  the  required 
tangent. 

116.    To  deduce  the  equation  of  the  7iormal  to  the  hyperbola^ 
We  can  do  this  by  operating  on  the  equation  of  the  tangent, 
as  in  previous  cases,  or  by  changing  h-  into  —  b'-^  in  the  equa- 
tion of  the  normal  to  the  ellipse.   Art.  86,   (3).     By  either 
method,  we  obtain 


y-y  =- 


cry 
1^ 


{x  -  x") 


(1) 


for  the  required  equation. 

117.    To  deduce  the  value  of  the  sub-normal. 

By  a  course  of  reasoning  similar  to  that  of  Art.  87,  we  have- 


sub-normal  =  — 5-  ^  ' 
a' 


CoK.    lib  =  a, 

sub-no7'mal  =  x" ; 
i.e.,  in  the  equilateral  hyperbola  the  sub-normal  is  equal  tO' 
the  abscissa  of  the  point  of  tangency. 


THE  HYPERBOLA.  153 


EXAMPLES. 

1.  Deduce  the  polar  equation  of  the  hyperbola,  the  pole 
being  at  the  centre. 

.2  ^ o^ 

a^  sin^  6  —  &^  cos  ^  6 

Write  the  equation  of  the  tangents  to  each  of  the  follow- 
ing hyperbolas,  and  give  the  value  of  the  sub-tangent  in 
each  case. 

2.  9  y'  -  4  a;2  =  —  36,    at    (4,  ord.  +). 

3.  yi  _  i^  =  _  1    at   (5,  ord.  +). 
9         16  '         V  .  -ry 

4.  ^  _  id  =  1    at    (4,  ord.  +). 

9        16  '         ^  '  ^ 

5.  2/^  --  4  a;2  =  —  36,    at    (abs.  +,  6). 

6.  ay'  —  bx^  =  —  ab,  at  (-Vab,  ord.  -|-)- 

7.  ^  -  ll  =  1,    at    (Vm,  0). 


8.  Write  the  equation  of  the  normal  to  each  of  the  above 
liyperbolas,  and  give  the  value  of  the  sub-normal  in  each 
case. 

9.  The  equation  of  a  chord  of  an  hyperbola  is  y  —  x  —  6 
=  0 ;  what  is  the  equation  of  the  supplemental  chord,  the 
axes  of  the  hyperbola  being  12  and.  8  ? 

Ans.     y  =  -X  —  -  . 
-^9         3 

10.  Given  the  equations 

^^  =  —  1    and  y  —  OS  =  0  : 

9  4  '  ^  ' 

required  the  equations  of  the  tangents  to  the  hyperbola  at  the 
points  in  which  the  line  pierces  the  curve. 


154  PLANE  ANALYTIC  GEOMETRY. 

11.  One  of  the  supplementary  chords  of  the  hyperbola 
9  2/^  —  16  rK^  =  —  144  is  parallel  to  the  line  y  =  x;  what  are 
the  equations  of  the  chords  ? 

Ans.      \  16  16. 

(-^9  3 

12.  Given  the  hyperbola  2  a;^  —  3  ^/^  =  6 ;  required  the 
equations  of  the  tangent  and  normal  at  the  positive  end  of 
the  right  hand  focal  ordinate. 

13.  What  is  the  equation  of  a  tangent  to 

^_  jT  _-, 
4  6    ~    ' 

which  is  parallel  to  the  line  2  y  —  a;  +  l  =  0? 

118.  The  angle  formed  by  the  focal  lines  drawn  to  any  point 
of  the  hyperbola  is  bisected  by  the  tangent  at  that  point. 

Making  ?/  =  o  in  the  equation  of  the  tangent  line,  Art. 
112,  (1),  we  have 

x  =  ^  =  OT.     Fig.  50. 
x' 

From  Art.  104,  (1)  OF  =  OF'  =  ae  ; 
hence  OF  -  OT  =  FT  =  ae  -  ^  =  ^  {ex"  -  a). 

OF'  +  OT  =  F'T  =  ae  +  -^  =  -^  (ex"  +  a); 

XX 

.:  FT  :  F'T  ::  ex"  -  a  :  ex"  +  a. 

But  from  Art.  105  we  have 

FP'  =  ex"  -  a 

FT'  =  ex"  +  a  ; 

. :  FF  :  FT' ::  ex"  -  a  :  ex"  +  a. 
Hence  FT  :  F'T  ::  FP' :  F'P' ; 

i.e.,  the  tangent  P'T  divides  the  base  of  the  triangle  FP'F' 
into  two  segments,  which  are  proportional  to  the  adjacent 
sides ;  it  must  therefore  bisect  the  angle  at  the  vertex. 


THE  HYPERBOLA.  155 

Cor.  Since  the  normal  P'N,  Fig.  50,  is  perpendicular  to 
the  tangent,  it  bisects  the  external  angle  formed  by  the  focal 
radii. 

ScHOL.  The  principle  of  this  article  gives  us  another 
method  of  drawing  a  tangent  to  the  hyperbola  at  a  given 
point.  Let  P'  be  the  point.  Fig.  50.  Draw  the  focal  radii 
FP',  F'P'.  The  line  P'T  drawn  so  as  to  bisect  the  angle 
between  the  focal  radii  will  be  tangent  to  the  curve  at  P'. 

119.  To  find  the  condition  that  the  line  y  =  sx  -\-  c  must 
fulfil  171  order  that  it  may  touch  the  hyperbola 

a""        b- 
By  a  method  similar  to  that  employed  in  Art.  89,  we  find 
s2  a''  -h''  =  (-'  .  .  .   (1) 
for  the  required  condition. 

Cob.  1.  Substituting  the  value  of  c  drawn  from  (1)  in  the 
equation  of  the  line,  we  have 


?/  =  sx  -1-  ^ s^a^  —  b'^    ...   (2) 

for  the  equation  of  the  tangent  to  the  hyperbola  in  terms  of  its 
slope. 

120.  To  find  the  locus  generated  by  the  intersection  of  a 
tangent  to  the  hyperbola  and  a  perpendicular  to  it  from  a  focus 
as  the  point  of  tangency  inoves  around  the  curve. 

x^+  tf^a''  .  .  .   (1) 

is  the  equation  of  the  required  locus.     (See  Art.  90.) 

121.  To  find  the  locus  generated  by  the  intersection  of  tivo 
tangents  ivhich  are  perpendicular  to  each  other  as  the  points  of 
tangency  move  around  the  curve. 

x^ -]- y^  =  a^  -  b''   .    .   .    (1) 

is  the  equation  of  the  required  locus.     (See  Art.  91.) 


156  PLANE  ANALYTIC   GEOMETRY. 

122.  Tivo  tangents  are  draivn  to  the  hyperbola  from,  a  point 
without ;  required  the  equation  of  the  line  joining  the  points  of 
tangency. 

^ ^ yy_  ^-^^         (i\ 

O  7  9  '        '         '  \       / 

a^  0- 

is  the  required  equation.     (See  Art.  92.) 

123.  To  find  the  equation  of  the  polar  of  the  pole  (x',  y'), 
with  regard  to  the  hypjerhola 

a^         b-'  ' 

^  ^  _  yy  __  -^         /-^ 

a^         b^ 
is  the  required  equation.     (See  Arts.  49  and  93.) 

124.  To  deduce  the  equation  of  the  hyperbola  tvhen  referred 
to  a  pair  of  conjugate  diameters. 

A  pair  of  diameters  are  said  to  be  conjugate  when  they  are  so 
related  that  the  equation  of  the  hyperbola,  when  the  cu7've  is 
referred  to  them  as  axes,  contains  only  the  second  p)owers  of  the 
variables. 

—  -  i^  =  1  ...   (1) 
a"'        b'^'  ^  ^ 

is  the  required  equation,  and 

ct^  sin  6  sin  cp  —  b"^  cos  6  cos  go  =  0, 

or  tan  6  tan  qo  =  —  ...   (2) 

a'^ 

is  the  condition  for  conjugate  diameters.     (See  Art.  94.) 

Cor.   Erom  the  form  of  (1)  we  see  that  all  chords  drawn 

parallel  to  one  of  two  conjugate  diameters  are  bisected  by  the  other. 

ScHOL.    From  Art.  Ill,  (1)  we  have 

b^ 

a^ 

hence  ss'  =  tan  d  tan  (jd. 

If,  therefore,  s  =  tan  6,  we  have  s'  =  tan  cp  ;  i.e.,  if  one  of 
tivo  conjugate  diameters  is  parallel  to  a  chord,  the  other  conjxi- 
gate  diameter  is  parallel  to  the  supplement  of  that  chord. 


THE  HYPERBOLA. 


157 


From  Art.  114  we  have 

a- 
hence  tt'  =  tan  6  tan  qp. 

If,  therefore,  t  =  tan  6,  we  have  t'  =  tan  q> ;  i.e.,  */  one  of 
two  conjugate  diametei's  is  parallel  to  a  tangent  of  the  liyper- 
bola,  the  other  conjugate  diameter  coincides  with  the  line  joining 
the  point  of  tangency  and  the  centre. 

125.    Prom  the  condition  for  conjugate  diameters, 


tan  6  tan  cp  = 


b"" 


we  see  that  the  products  of  the  slopes  of  any  pair  of  conju- 
gate diameters  is  positive ;  hence,  the  slopes  are  both  positive 
or  both  negative.  It  appears,  therefore,  that  any  two  conju- 
gate diameters  must  lie  in  the  same  quadrant. 

126.    To  find  the  equation  of  a  conjugate  diameter. 


Fig.  51. 


158  PLANE  ANALYTIC  GEOMETRY. 

Let  V'R"  be  any  diameter ;  then  P'E',  drawn  through  the 
centre  0  parallel  to  the  tangent  at  P"  (P^N')  will  be  its  eon- 
jugate  diameter.     Art.  124,  Schol. 

The  equation  of  the  tangent  at  P"  (x",  y")  is 

•   (1) 


XX         yy    _^ 

a?            V^ 

hence,  the 

equation  of  P'R'  is 

^^"      yy"  __  0 

or 

V'      x" 
y  =  ^r-  —TT^    •   ■ 

«*•    y 

But 

^  =  cot  P"OX  = 

1 

y 

s 

hence 

y  =  -^x  .  .  .  (£ 

') 

(2) 


as 

is  the  equation  of  a  diameter  in  terms  of  the  slope  of  its  conju- 
gate diameter. 

127.  To  find  the  co-ordinates  of  either  extremity  of  a 
diameter,  the  co-ordinates  of  one  extremity  of  its  co7ijugate 
diameter  being  given. 

Let  the  co-ordinates  of  P"  (x",  y"),  Fig.  61,  be  given. 

By  a  course  of  reasoning  similar  to  that  of  Art.  96,  we  find 

«=-[--  y",  rf  =  4--X  . 
0  a 

The  upper  signs  correspond  to  the  point  P'  {x',  y')  ;  the 
loAver  signs  to  the  point  R'  (—  x',  —  y'). 

128.  To  shoiv  that  the  difference  of  the  squares  of  any  pair 
of  semi-conjugate  diameters  is  equal  to  the  difference  of  the 
squares  of  the  semi-axes. 

By  a  course  of  reasoning  similar  to  that  of  Art.  97,  or,  by 
substituting  —  b^  for  b^,  —  b'^  for  b'^  in  (4)  of  that  article,  we 

find 

a'2  _  S'2  _  ^2  _  52  _  _   _   ^ij 


THE  HYPERBOLA.  159 

Cor.  If  a  =  b,  then  a'  =b' ;  i.e.,  the  equilateral  hyperbola 
has  equal  conjugate  diameters. 

129.  To  show  that  the  'parallelogram  constructed  on  any  two 
conjugate  diameters   is  equivalent  to  the  rectangle  constructed 

on  the  axes. 

By  a  method  similar  to  that  of  Art.  98,  we  can  show  that 

4  a'b'  sin  (go  -  ^)  =  4  ab ; 
i.e.,  Area  MNM'N'  =  Area  CDC'D'.    Fig.  51. 

EXAMPLES. 

1.  The  line  y  =  2x  -\-c  touches  the  hyperbola 

x^       y^ 1 

what  is  the  value  of  c  .^  

Ans.     c  =  i  V32. 

2.  A  tangent  to  the  hyperbola 

x^  _   ?/^   _  -1 
10        12  ~ 
has  its  Y-intercept  =  2 ;  required  its  slope  and  equation. 

Ans.     VlTe  ;  y  =  ViTG  a;  +  2. 

3.  A  tangent  to  the  hyperbola  4?/^  —  2x^  =  6  makes  an 
angle  of  45°  with  the  X-axis  ;  required  its  equation. 

4.  Two  tangents  are  drawn  to  the  hyperbola  A  y^  —  2  x^  ^ 
—  36  from  the  point  (1,  2)  ;  required  the  equation  of  the  chord 
of  contact. 

Ans.  9  cc  —  8  y  ^  36. 

5.  What  is  the  equation  of  the  polar  of  the  right-hand 
focus  ?     Of  the  left-hand  focus  ? 

6.  What  is  the  polar  of  (1,  ^)  with  regard  to  the  hyperbola 

4y^  —  x'^  =  —  4?  Ans.     x  —  2  y  =  4. 

7.  Find  the  diameter  conjugate  to  y  =  x  in  the  hyperbola 

-^  _  i^  =  1 
9         16 

Ans.     y  =  ^  X. 


160 


PLANE  ANALYTIC  GEOMETRY. 


8.  Given  the  chord  y  —  2x  -\-  Q  oi  the  hyperbola 

required  the  equations  of  the  supplementary  chord. 

Ans.     y  =  ^  X  —  §. 

9.  In  the  last  example  find  the  equation  of  the  pair  of 
conjugate  diameters  which  are  parallel  to  the  chords. 

Ans.     y  =  2x,  9  y  =  2x. 

10.  The  point  (5,  ^^)  lies  on  the  hyperbola  9y-  —  IQx^  = 
—  144;  required  the  equation  of  the  diameter  passing  through 
it;  also  the  co-ordinates  of  the  extremities  of  its  conjugate 
diameter. 

130.  To  deduce  the  equations  of  the  rectilinear  asymptotes 
of  the  hyperbola. 

An  Asymptote  of  a  curve  is  a  line  passing  within  a  finite 
distance  of  the  origiii  which  the  curve  continually  approaches^ 
and  to  which  it  becomes  tangent  at  an  infinite  distance. 


Fig.  52. 


THE  HYPERBOLA.  161 

The  equation  of  the  hyperbola  whose  transverse  axis  lies 
along  the  X-axis  may  be  put  under  the  form 

The  equations  of  the  diagonals,  DD',  CC,  of  the  rectangle 
constructed  on  the  axes  AA',  BB'  are 

y'  =  ^-x, 
a 

or,  squaring,      y''^  =  ^x^  .  .  .  (2) 

where  y'  represents  the  ordinates  of  points  on  the  diagonals. 
Let  P'  {x,  y)  be  any  point  on  the  X-hyperbola ;  and  let  D" 
(x,  y')  be  the  corresponding  point  on  the  diagonal  DD'.     Sub- 
tracting (1)  from  (2)  and  factoring,  we  have 

hence  /  -  2/  =  ^"^'  =     /'        •  •  •   (3) 

As  the  points  D",  P'  recede  from  the  centre,  0,  their  ordi- 
nates D"]Sr,  P'N  increase  and  become  infinite  in  value  when 
D"  and  P'  are  at  an  infinite  distance.  But  as  the  ordinates 
increase  the  value  of  the  fraction  (3),  which  represents  their 
difference,  decreases  and  becomes  zero  when  y'  and  y  are 
infinite  ;  hence,  the  points  D''  and  P'  are  continually  approach- 
ing each  other  as  they  recede  from  the  centre  until  at  infinity 
they  coincide.  But  the  locus  of  D"  during  this  motion  is  the 
infinite  diagonal  DD' ;  hence,  the  diagonals  of  the  rectangle 
constructed  on  the  axes  of  the  hyperbola  are  tha  asymptotes  of 
the  curve. 

Therefore        y  z=  -\ —  x  and  tj  = x 

a  a 

are  the  reqiiired  equations. 

CoR.  1.    If  a-  =  h,  then 

y  =  -\-  X  and  y  =.  —  x\ 

i.e.,  the  asymptotes  of  the  equilateral  hyperbola  make  angles  of 

45°  with  the  IL-axis. 


162 


PLANE  ANALYTIC  GEOMETRY. 


Cor.  2.    The   equation  of   the  hyperbola  conjugate  to  (1) 
may  be  put  under  the  form 


^.2 


-  (x'^  +  a^)  .  .  .  (4) 


Subtracting  (1)  from  (4),  we  have 


y 


y  =  PiP'  = 


2  b'' 


y"  +  y 

hence,  an  hyperbola  and  its   conjugate  are   curvilinear  asymp- 
totes of  each  other. 

CoK.  3.    Subtracting  (2)  from  (4),  we  have 

b- 


y 


y'  =  P,D"  = 


y"^y" 

hence,  the  rectilinear  asymptotes  of  an  hyperbola  and  of  its  con- 
jugate are  the  same. 

131.    To  deduce  the  equation  of  the  hyperbola  when  referred 
to  its  rectilinear  asym,ptotes  as  axes. 

Y 


Fig.  53. 


The  equation  of  the  hyperbola  when  referred  to  OY,  OX, 


IS 


1  . 


(1) 


THE  HYPERBOLA.  163 

We   wish  to   ascertain   what  this  equation  becomes  when 
OY',  OX'  the  rectilinear  asymptotes  are  taken  as  axes. 

Let  P'  be  any  point  of  the  curve ;  let  Y'OX  =  XOX'  =  6. 
Then  (OC,  OF)  =  {x,  y) ;  (OD,  DP')  =  {x',  ij'). 

Prom  the  figure,  OC  =  OK  +  DP ;  CP'  =  RP'  -  DK ; 
i.e.,  x  =  {x'  +  y')  cos  6  \  y  =  {y  —  x')  sin  Q. 

But  from  the  triangle  OAB,  we  have 
AB  h 


sin  B 


cos  it  = 


OB         ^ci?  +  ^2 
OA  a 


OB     Va-  +  y" 


hence,  x  =  ix'  -{-  y')  — ==:^^- ;  y  =  iv'  —  ^')  — :  . 

Substituting  these  values  in  (1),  we  have 

(^'  +  yj  -  in'  -  xy  =  «'  +  ^^ 

or,  reducing  and  dropping  accents, 

xy  =  — J—  ...   (2) 

for  the  equation  of  the  hyperbola  referred  to  its  asymptotes. 
In  a  similar  manner  we  may  show  that 

xy  = ^  ■  •  •   (3) 

is  the  equation  of  the  hyperbola  conjugate   to   (1),  when  re- 
ferred to  its  asymptotes  as  axes. 

Cor.    Multiplying  (2)  by  sin  2  6  we  may  place  the  result  in 
the  form 


vx  sin  2  ^  =  Va^_+1' .  ^/^!_±i!  siu  2  ^  • 

2  2  ' 

that  is  DP'.  P'H'  =  ON.  AH ; 

therefore  area  ODP'P  =  area  OMAN ; 

hence,  the  area  of  the  parallelogravi  constructed  upon  the  co- 
ordinates of  any  point  of  the  hyperbola,  the  asymptotes  being 
axes,  is  constant  and  equal  to  the  area  of  the  rhombus  con- 
structed upon  the  co-ordinates  of  the  vertex. 


164 


PLANE  ANAL YTIC   GEOME TR Y. 


132.    To  deduce  the  equation  of  the  tangent  to  the  hyperbola 
when  the  curve  is  referred  to  its  rectilinear  asymptotes  as  axes. 


Fig.  54. 


By  a  course  of  reasoning  similar  to  that  employed  in  ArtSo 
41,  57,  82,  we  find  the  required  equation  to  be 


y  —  y 


or,  symmetrically. 


x"  ^  y" 


y^^x-  X") 


■  ■   (2) 


(1) 


CoR.    If  we  make  ?/  =  0  in  (2),  we  have 

x  =  2x"  =  OT.     Fig.  54. 
But  OM  =  x",  .:  OM  =  MT  .-.  T'D  =  TD ; 

hence,  the  point  of  tangency  in  the  hyperbola  bisects  that  por- 
tion of  the  tangent  included  between  the  asymptotes. 


THE  HYPERBOLA.  165 

133.  Since  D  (x",  y")  is  a  point  of  the  hyperbola,  we  have 
(see  Fig.  54) 

4  x"y"  =a^  +  b% 

or  2x"  .2  y"  =  a-  +  h"- ; 

i.e.,  OT  .  OT'  =  a'  ^y  .  .  .  (1) 

hence,  the  rectangle  of  the  intercepts  of  a  tangent  on  the  asymp- 
totes is  constant  and  equal  to  the  sum  of  the  squares  on  the 
semi-axes. 

134.  From  (1)  of  the  last  article  we  have,  after  multiply- 

,1  ,   ,       sin  2  6 

mg  through  by  — - —  , 

^'^  •  ^^'  sin  2  ^  =  ^J—±Jl.  sin  2  ^  =  {a^  +  h'-)  sin  9  cos  6. 

But,  Art.  131, 

sm  ii  =  —  ,  cos  9  = 


hence sm  2  9  ^  ab  ; 

i.e.,  area  OTT'  =  area  OAD'B. 

.-.  the  triangle  formed  by  a  tangent  to  the  hyperbola   and  its 
asymptotes  is  equivalent  to  the  rectangle  on  the  serai-axes. 

135.    Draw  the  chord  RE,',  Fig.  54,  parallel  to  the  tangent 
T'T.     Draw  also  the  diameter  OL  through  D. 
Since  TD  =  T'D,  we  have  P/L  =  RL. 
Since  OL  is  a  diameter,  we  have  LK  =  LH ;  hence 

R'L  —  LK  =  RL  -  LH ; 

i.e.,  R'K  =  RH  ; 

hence,  the  intercepts  of  a  chord  between  the  hyperbola  and  its 
asymptotes  cure  equal. 


166  PLANE  ANALYTIC  GEOMETRY. 

EXAMPLES. 
1.    What  are  the  equations  of  the  asymptotes  of  the  hyper- 
bola ^  _  i^  =  1? 
9        16 

Ans.     y  =  _1_  |-  cc. 
What  are  the  equations  of  the  asymptotes  of  the  follow- 
ing hyperbolas : 


2. 

16       ^ 

Ans. 

,     X 

4. 

10 

~f-' 

3. 

3  2/2  _  2  a;2  = 

-6. 

5. 

mx^  - 

-  ny^  = 

Ans.     y  ^  i  Vf  X. 

6.  What  do  the  equations  given  in  the  four  preceding  ex- 
amples become  when  the  hyperbolas  which  they  represent  are 
referred  to  their  asymptotes  as  axes  ? 

7.  The  semi-conjugate  axis  of  the  hyperbola  xy  =  25  is 
6 ;  what  is  the  value  of  the  semi-transverse  axis  ? 

A71S.     8. 

What  are  the  equations  of  the  tangents  to  the  following 
hyperbolas : 

8.  To    xy  =  10,  at  (1,  10). 

Ans.     y  -\-10x  =  20. 

9.  To    xy  =  +  12,  at  (2,  6). 

Ans.     y  =  —  3x  +  12. 

10.  To    xy  =  m,  at  (—  1,  —m). 


11.    To    xy=-p,  at  (-2,|^ 


12.  Required  the  point  of  the  hyperbola  xy  =  12  for  which 
the  sub-tangent  =  4. 

Ans.     (4,  3). 

13.  The  equations  of  the  asymptotes  of  an  hj^perbola 
whose  transverse  axis  =  16  are  3  ?/  =  2  x  and  3  ?/  +  2  ic  =  0 ; 
required  the  equation  of  the  hyperbola. 

Ans.     ^_^  =  i, 
64       256 


THE   HYPERBOLA.  167 

14.  Prove  that  the  product  of  the  perpendiculars  let  fall 
from  any  point  of  the  hyperbola  on  the  asymptotes  is  con- 
stant and 

«2  -I-  &2  • 

GENERAL   EXAMPLES. 

1.  The  point  (6,  4)  is  on  the  hyperbola  whose  transverse  is 
10 ;  required  the  equation  of  the  hyperbola. 

Ans. ^  =  1. 

25        400 

2.  Assume  the  equation  of  the  hyperbola,  and  show  that 
the  difference  of  the  distances  of  any  point  on  it  from  the 
foci  is  constant  and  =  2  a. 

3.  Required  the  equation  of  the  hyperbola,  transverse 
axis  =  6,  which  has  5  y  =  2  x  and  3  ?/  =  13  a;  for  the  equa- 
tions of  a  pair  of  conjugate  diameters. 


X 


btf_ 


Ans.     _  -  ^1^  =  1. 
9         78 

4.  Show  that  the  ratio  of  the  sum  of  the  focal  radii  of  any 
point  on  the  hyperbola  to  the  abscissa  of  the  point  is  con- 
stant and  =  2  e. 

5.  What  are  the  conditions  that  the  line  y  =:  sx  -\-  c  must 
fulfil  in  order  to  touch 

—  —  -^  =  1  at  infinity  ? 

Ans.     s  =  -J-  -  ,  c  =  0. 
a 

6.  Show  that  the  conjugate  diameters  of  an  hyperbola  are 
also  the  conjugate  diameters  of  the  conjugate  hyperbola. 

7.  Show  that  the  portions  of  the  chord  of  an  hyperbola 
included  between  the  hyperbola  and  its  conjugate  are  equal. 

8.  What  is  the  equation  of  the  line  which  passes  through 
the  focus  of  an  hyperbola  and  the  focus  of  its  conjugate 
hyperbola  ? 

Ans.     X  -\-  y  ^  -y/ a^  -(-  b^. 


168  PLANE  ANALYTIC  GEOMETRY. 

9.  Show  that 

e'       a 
when  e  and  e'  are  the  eccentricities  of  two  conjugate  hyper- 
bolas.     } 

10.  Find  the  angle  between  any  pair  of  conjugate  diame- 
ters of  the  hyperbola. 

11.  Show  that  in  the  hyperbola  the  curve  can  be  cut  by 
only  one  of  two  conjugate  diameters. 

12.  Find  whether  the  line  y  =  ^x  intersects  the  hyperbola 
16  y-  —  9  a;-  —  —  144,  or  its  conjugate. 

13.  Show  that  the  conjugate  diameters  of  the  equilateral 
hyperbola  make  equal  angles  with  the  asymptotes. 

14.  Show  that  lines  drawn  from  any  point  of  the  equilat- 
eral hyperbola  to  the  extremities  of  a  diameter  make  equal 
angles  with  the  asymptotes. 

15.  In  the  equilateral  hyperbola  focal  chords  drawn  parallel 
to  conjugate  diameters  are  equal, 

16.  A  perpendicular  is  drawn  from  the  focus  of  an  hyper- 
bola to  the  asymptote  :  show 

(a)  that  the  foot  of  the  perpendicular  is  at  the  distance  a 
from  the  centre,  and 

{b)  that  the  foot  of  the  perpendicular  is  at  the  distance  h 
from  the  focus. 

17.  For  what  point  of  an  hyperbola  is  the  sub-tangent  = 
the  sub-normal  ? 

18.  Show  that  in  the  equilateral  hyperbola  the  length  of 
the  normal  is  equal  to  the  distance  of  the  point  of  contact 
from  the  centre. 

19.  Show  that  the  tangents  drawn  at  the  extremities  of  any 
chord  of  the  hyperbola  intersect  on  the  diameter  which 
bisects  the  chord. 


THE  HYPERBOLA.  169 

20.  Find  the  equation  of  the  chord  of  the  hyperbola 

^  —  J^  =1 
9         12 

which  is  bisected  at  the  point  (4,  2). 

21.  Eequired  the  equations  of  the  tangents  to 

16        10 
which  make  angles  of  60°  with  the  X-axis. 

22.  Show  that  the  rectangle  of  the  distances  intercepted  on 
the  tangents  drawn  at  the  vertices  of  an  hyperbola  by  a 
tangent  drawn  at  any  point  is  constant  and  equal  to  the 
square  of  the  semi-conjugate  axis. 

23.  Given  the  base  of  a  triangle  and  the  difference  of  the 
tangents  of  the  base  angles  ;  required  the  locus  of  the  vertex, 

24.  Show  that  the  polars  of  (?».,  n)  with  respect  to  the 
hyperbolas 

^  —  l!-  =  1,     ll  —  ^  =  1  are  parallel. 

25.  If  from  the  foot  of  the  ordinate  of  a  point  (ic,  y)  of  the 
hyperbola  a  tangent  be  drawn  to  the  circle  constructed  on 
the  transverse  axis,  and  from  the  point  of  tangency  a  line  be 
drawn  to  the  centre,  the  angle  which  this  line  forms  with  the 
transverse  axis  is  called  the  eccentric  angle  of  (a;,  y).  Show 
that  (cc,  ?/)  =  {a  sec  qo,  b  tan  qo),  and  from  these  values  deduce 
the  equation  of  the  hyperbola. 

26.  If  («',  y'),  {x",  y")  are  the  extremities  of  a  pair  of 
conjugate  diameters  whose  eccentric  angles  are  qo'  and  go,  show 
that  (p'  +  9  =  90°. 


170  PLANE  ANALYTIC  GEOMETRY. 


CHAPTER   IX. 
THE   GENERAL   EQUATION   OF  THE    SECOND    DEGREE. 

136.  The  most  general  equation  of  the  second  degree  be- 
tween two  variables  is 

ay~  +  bxy  -\-  cx^  -{-  dy  -\-  ex  +  f  ==  0  .   .  .   (1) 

in  which  a,  b,  c,  d,  e,  f  are  any  constant  quantities  whatever. 
To  investigate  the  properties  of  the  loci  which  this  equation 
represents  under  all  possible  values  of  the  constants  as  to 
sign  and  magnitude  is  the  object  of  this  chapter. 

137.  The  equations  of  the  lines  in  a  plane,  with  which  we 
have  had  to  do  in  preceding  chapters,  are 

Ax  +  By  +  C  =  0.     Straight  line. 

(Ace  +  B?/  4"  C)^  =  0.     Two  coincident  straight  lines. 

2/2  _  aj2  _  Q_     rji^^Q  straight  lines. 

2/2  -|-  a;^  =  a".     Circle. 

?/2  -j-  a;2  =  0.     Two  imaginary  straight  lines. 

y"^  =  2,  px.     Parabola. 

aY  +  h^'x^  =  a%\     Ellipse. 

o?y^  —  b^x^  =  —  a^b^.     Hyperbola. 

a^y-  —  b'^x"  —  a-b^.     Hyperbola. 

Comparing  these  equations  with  the  general  equation,  we 
see  that  all  of  them  may  be  deduced  from  it  by  making  the 
constants  fulfil  certain  conditions  as  to  sign  and  magnitude. 
We  are,  therefore,  prepared  to  expect  that  the  lines  which 
these  equations  represent  will  appear  among  the  loci  repre- 
sented by  the  general  equation  of  the  second  degree  between 
two  variables.  In  the  discussion  which  is  to  ensue  we  shall 
find  that  these  lines  are  the  only  loci  represented  by  this 
equation. 


EQUATION  OF  THE   SECOND  DEGREE.  171 

DISCUSSION. 

138.  To  show  that  the  locus  represented  by  a  coynjplete  equa- 
tion of  the  secoiid  degree  hetxoeen  two  variables  is  also  represented 
by  an  equation  of  the  second  degree  betiveen  two  variables,  in 
which  the  term  containing  xy  is  wanting. 

Let  us  assume  the  equation 

ay"^  +  ^^y  +  ^^^  +^y  +  ^^  +  /  =  0  •  •  •  (1) 
and.  refer  the  locus  it  represents  to  rectangular  axes,  making 
the  angle  6  with  the  old  axes,  the  origin  remaining  the  same. 
From  Art.  33,  Cor.  2,  we  have 

X  =  x'  cos  9  —  y'  sin  0 

y  =  x  sin  6  -{-  y'  cos  6 
for  the  equations  of  transformation.    Substituting  these  values 
in  (1),  we  have, 

aY'  +  b'x'y'  +  c'x'^  +  d'y  +  e'x'  +/  =  0  .  .  .   (2) 
in  which 

a'  =  a  cos^  6  -\-  c  sin^  9  —  b  sin  9  cos  9 

b'  =  2  {a—  c)  sin  ^  cos  ^  +  ^»  (cos^  9  —  sin^  9) 

c'  =  a  sin2  9  ^  c  cos^  ^  +  5  sin  ^  cos  ^  J.    ...   (3) 

d'  ^  d  cos  9  —  e  sin  9 

e'  =^  d  sin  9  -\-  e  cos  9 

Since  9,  the  angle  through  which  the  axes  have  been  turned, 
is  entirely  arbitrary,  we  are  at  liberty  to  give  it  such  a  value 
as  will  render  the  value  of  b'  equal  to  zero.  Supposing  it  to 
have  that  value,  we  have 

2  (a  —  c)  sin  9  cos  9  +  b  (cos^  9  —  sin^  $)  =  0, 
or  (a  —  c)  sm  2  ^  +  6  cos  2  ^  =  0  .  .  .   (4) 

or  tan  29  =  -^—  ...  (5) 

c  —  a 

Since  any  real  number  between  -|-  oo  and  —  go  is  the  tan- 
gent of  some  angle,  equation  (5)  will  always  give  real  value 
for  2  9  ;  hence  the  above  transformation  is  always  possible. 
Making  &'  =  0  in  (2),  we  have,  dropping  accents, 

a'y''  +  cV  ^d'y  ■\-e'x-^f=^  .  .  .   (6) 
for  the   equation  of  the  locus  represented  by  (1).     To   this 
equation,  then,  we  shall  confine  our  attention. 


172  PLANE  ANALYTIC  GEOMETRY. 

CoE.  1.  To  find  the  value  of  a'  and  c'  iii  terms  of  a,  b, 
and  c.  Adding  and  then  subtracting  the  first  and  third  of 
the  equations  in  (3),  we  have 

c'  -\-a'  =  c  -\-  a  .  .  .   {!) 
c'  —  a'  =  (c  —a)coB2  6  -\-bB\n26  .  .  .   (8) 

Squaring  (4)  and  adding  to  the  square  of  (8),  we  have 
(c'  -  a'y  =  (c  -  ay  +  b^ ; 


.-.  c'  —  a'  =  V(c  -  af  -\-b^  .  .  .   (9) 
Subtracting  and  then  adding  (7)  and  (9),  we  have 


a'  =  i\c  +  a-  V(c  -  af  -{-  b^   .  .  .   (10) 
c'  =  1  ^c  +  a  +  V(c  -  ay  +  b'l   .  .  .   (11) 

Cor.  2.  To  find  the  signs  of  a'  and  c'.  Multiplying  (10) 
and  (11),  we  have 

a'c'  =  \\{c  +  ay  -  {{c-ay  -^b')\; 
.:  a'c'  =  -i{b^  -4.ac)  .  .  .   (12) 

Hence,  the  sigiis  of  a'  and  c'  depend  upon  the  sign  of  the 
quantity  P  —  4:  ac. 

The  following  cases  present  themselves  : 

1.  b^  <C.  4:  ac.  The  sign  of  the  second  member  of  (12)  is 
positive,  ,-.  a'  and  c'  are  both  positive,  or  both  ?iegative. 

2.  b"^  =  4zac.  The  second  member  of  (12)  becomes  zero,  .-. 
a'  =  0,  or  c'  =  0. 

[It  will  be  observed  that  a'  and  c'  cannot  be  equal  to  zero 
at  the  same  time,  for  such  a  supposition  would  reduce  (6)  to 
an  equation  of  the  first  degree.] 

3.  6^  >  4  ac.  The  sign  of  the  second  member  of  (12)  is 
negative,  .-.  a'  must  be  positive  and  c'  negative,  or  a'  must  be 
negative  and  c'  positive. 

139.  To  transform  the  equation  a't/"^  -\-  c'x^  +  d'y  -\-  e'x  -\- 
y  =  0  into  an  equation  in  which  the  first  pmwers  of  the  vari- 
ables are  missing. 


EQUATION  OF  THE  SECOND  DEGREE.  173 

Let  us  refer  the  locus  to  a  parallel  system  of  rectangular 
axes,  the  origin  being  at  the  point  (m,  n).  From  Art.  32,  we 
have 

X  =  VI  -\-  x',  y  ^=  n  -{-  y'. 

Substituting  these  values  in  the  given  equation,  we  have 
a'y'2  +  c'x""  +  d"y'+  e"x'+f"  =  0  ...   (2) 
in  which 

d"  =2a'n  +  d'  1 

6"  =  2  c'm  +  e'  !-    .  .  .   (3) 

f"  =  a'n^  +  c'lv?  -\-  d'n  +  e'ln  +  /  J 
Since  m  and  n  are  entirely  arbitrary,  we  may,  in  general, 
give  them  such  values  as  to  make 

2  a'?i  +  cZ'  =  0  and  2  c'm  +  e'  =  0 ; 
i.e.,  in  general,  we  may  make 

and  m  =  -  -^  ...  (4) 


2  a'  2  c' 

We  see  from  these  values  that  when  a'  and  c'  are  not  zero, 
this  transformation  also  is  possible  ;  and  equation  (2)  becomes, 
after  dropping  accents, 

ay+flV+r  =  0  ...  (5) 

Equation  (5),  we  observe,  contains  only  the  second  power  of 
the  variables ;  hence  it  is  satisfied  for  the  points  (x,  y)  and 
(—  X,  —  y).  But  only  the  equation  of  curves  with  centres 
can  satisfy  this  condition ;  hence,  equation  (5)  is  the  equa- 
tion of  central  loci.  When  either  a'  or  c'  is  zero,  then  n  or  m 
is  infinite  and  the  transformation  becomes  impossible.  Hence 
arise  two  cases  which  require  special  consideration. 

140.    Case   1.    a'  =  o. 

Under  this  supposition  equation  (6),  Art.  138,  becomes 

c'x-  +  d'y  +  e'x  +/=  0  ...   (1) 
Referring  the  locus  of  this  equation  to  parallel  axes,  the 
origin  being  changed,  we  have   for  the   equations    of   trans- 
formation 

X  =  m  -\-  x',  y  =  n  -\-  y'. 


174  PLANE  ANALYTIC  GEOMETRY. 

Substituting  in  (1),  we  have 
c'x'^  +  d'y'  +  (2  dm  +  e')  x'  +  c'm''  +  d'n  -\- e'm -\-f  =0  .  .  .  (2) 

Kow,  in  general,  we  may  give  vi  and  ?i  such  values  as  to 
make 

2  c'??i  -[-  e'  =  0,  and  c'm^  +  c?'?i  +  e'm  -|-y  =  0 ; 
i.e.,  we  may  make 


m  = ,  and 


e 
c'm^  +  e^«?,  +  /  _  e''  -  4/c^ 


;.    .  .  .   (a) 


If  c^'  is  not  zero  (since  a'  =  0,  c'  is  not  zero),  this  transfor- 
mation is  possible  and  (2)  becomes,  after  dropping  accents, 
c'x^  +  d^y  =  0, 

°^  d' 

x^=-^y  .  .  .   (3) 

CoR.    If  d'  =  0,  (1)  becomes 

cV-f  e'a;-f /  =  0  ...   (4) 
or,  solving  with  respect  to  x, 


2  c'  ^  ^ 

141.    Case  2.    c'  =  o. 

Under  this  supposition  equation  (6),  Art.  138,  becomes 
ay  +  d'y  +  e'x+f=0  ...   (1) 

Transforming  this  equation  so  as  to  eliminate  y  and  the 
constant  term,  by  a  method  exactly  similar  to  that  of  the 
preceding  article,  we  find 

d' 

n  = , 

2  a'  ' 

d'^  -  4  a'f 

4  a'e'      ' 

and,  if  e'  is  not  zero,  we  have  (a'  is  not  zero  since  c'  =  0) 

f=-^,x  .  .  .  (2) 
a 


EQUATION  OF  THE  SECOND  DEGREE. 


175 


Cob.    If  e'  =  0,  equation  (1)  becomes 


y 


d'  ^  Vd'^  -  4.  fa' 


(3) 


142.    Summarizing  the  results  of  the  preceding  articles,  we 
find  that  the  discussion  of  the  general  equation 

ay^  -j-  hxi/  -^  cx'^  ->r  di/  -{-  ex  +  f  =  0 
has  been  reduced  to  the  discussion  of  the  three  simple  forms : 

1.    a' if  +  c'x-  -\-f"  =  0.     Art.  139,  (5) 

f=-I.tj.     Art.  140,  (3) 


2/2  =  _  1-  ic.     Art.  141,  (2) 

_  e'  jz  ^e'-'  -  ^fo' 
^~  2  c' 


-d'  :^  ^d'^  -  4.  fa' 
2  a' 


Art.  140,  (5) 
Art.  141,  (3) 


The  discussion  now  involves  merely  a  consideration  of  the 
sign  and  magnitude  of  the  constants  which  enter  into  these 
equations. 

143.  b^  <'i  ac. 

Under  this  supposition,  since  a'  and  c'  are  both  positive  or 
both  negative,  Art.  138,  Cor.  2,  neither  a'  nor  c'  can  be  zero ; 
hence,  forms  2  and  3  of  the  preceding  article  are  excluded 
from  consideration. 

The  first  form  becomes  either 

ay  +  c'x^+/"=0,      1  .^^ 

or  _ay-cV+.r=0| 

in  which  a'  and  c'  may  have  any  real  value  and  /'  may  have 
any  sign  and  any  value.     Hence  arise  four  cases : 


176 


PLANE  ANALYTIC  GEOMETRY. 


Case  1.    If  f"  has  a  sign  different  from  that  of  a'  and  c', 
equations   (1)   are  equations  of  ellipses  whose  semi-axes  are 


v/^-dj=Y//; 


Case  2.  If  /"  has  the  same  sign  as  that  of  a'  and  c',  equa- 
tions (1)  represent  imaginary  curves. 

Case  3.  If  a'  =  c'  and/"  has  a  different  sign  from  that  of 
a'  and  c',  equations  (1)  are  equations  of  circles.  If  f"  has 
the  same  sign  as  a'  and  c',  then  the  equations  represent  imagi- 
nary curves. 

Case  4.  If/"  =  0,  equations  (1)  are  equations  of  t^vo  imagi- 
nary  straight  lines  jmssing  through  the  origin. 

Hence,  when  5^  <  4  ac,  every  equation  of  the  second  degree 
bettveen  two  variables  represents  an  ellipse,  an  imaginary  curve, 
a  circle,  or  two  imaginary  straight  lines  intersecting  at  the 
origin. 

144.  ^2  =  4  ac. 

Under  this  supposition,  Art.  138,  Cor.  2,  either  a'  =  0,  or 
c'  =  0 ;  hence,  form  (1)  of  Art.  142  is  excluded. 

Resuming  the  forms 


y  =  - 


d' 
e' 


(2) 


_  e'  _J_  Ve'-  -  4  fc' 
"  = 2^^ -~ 


y 


^  -d:  ^^  ^d'^-  -  ^fa' 

~  2  a' 


(3) 


J 


we  have  four  cases  depending  upon  the  sign  and  magnitude 
of  the  constants. 

Case  1.  If  d'  and  c'  in  the  first  form  of  (2)  are  not  zero, 
and  if  e'  and  a'  in  the  second  form  of  (2)  are  not  zero,  then 
equations  (2)  are  equations  of  parabolas. 


EQUATION  OF   THE   SECOND  DEGREE.  177 

Case  2.  Since  the  first  form  of  (3)  is  independent  of  y,  it 
represents  tiuo  lines  parallel  to  each  other  and  to  the  Y-axis. 
The  second  form  of  (3)  represents,  similarly,  tivo  li7ies  which 
are  parallel  to  the  X-axis. 

Case  3.  If  e"^  <  4/c'  the  first  form  of  (3)  represents  tivo 
imaginary  lines. 

If  d'"^  <  4:  fa',  the  second  form  of  (3)  represents  tivo  imagi- 
nary lines. 

Case  4.  If  e'^  =  4/c',  the  first  form  of  (3)  represents  one 
straight  line  parallel  to  the  Y-axis. 

If  c^'2  _  4.faf,  the  second  form  of  (3)  represents  one  straight 
line  parallel  to  the  X-axis. 

Hence,  when  &^  :=  4  ac,  every  equation  of  the  second  degree 
between  two  variables  represents  a  parabola,  two  parallel  straight 
lines,  two  hnaginary  lines,  or  one  straight  line, 

145.  b''>  4.  ac. 

Under  this  supposition,  Art.  138,  Cor.  2,  since  a'  and 
c'  must  have  opposite  signs,  neither  a'  nor  c'  can  be  zero ; 
hence  forms  (2)  and  (3)  of  Art.  142  are  excluded  from  con- 
sideration under  this  head.     The  first  form  becomes  either 

ay-cV+/"  =  0       1 
or  -aY  +  c'x'^f"  =0]    "   "  "   (^) 

We  have  here  three  cases. 

Case  1.  If  f  has  a  different  sign  from  that  of  a',  equations 
(1)  are  equations  of  hyperbolas  whose  semi-axes  are 


«  =  y//;andi=y'r. 


If  /"  has  a  different  sign  from  that  of  c',  equations  (1)  are 
still  equations  of  hyperbolas. 

Case  2.  If  a'  =  c' ,  equations  (1)  are  equations  of  equilat- 
eral hyperbolas. 

Case  3.  If  f"  =  0,  equations  (1)  are  equations  of  two  inter- 
secting straight  lines. 


178  PLANE  ANALYTIC   GEOMETRY. 

Hence,  when  6^  >  4  ac,  every  equation  of  the  second  degree 
between  two  variables  represents  an  hyperbola,  an  equilateral 
hyperbola,  or  two  intersecting  straight  lines. 

146.  Summary.  The  preceding  discnssion  has  elicited  the 
following  facts  : 

1.  That  the  general  equation  of  the  second  degree  between 
tivo  variables  represents,  under  every  conceivable  value  of  the 
constants  which  enter  into  it,  an  ellipse,  a  parabola,  an  hyper- 
bola, or  one  of  their  limiting  cases. 

2.  When  b'^  <  4ac  it  represents  an  ellipse,  or  a  limiting  case. 

3.  When  h'^  =  4:ac  it  represents  a  parabola,  or  a  limiting 
case. 

4.  When  b'^  >  4ac  it  represents  an  hyperbola,  or  a  limiting 
case. 

EXAMPLES. 

1.  Given  the  equation  3  2/^  +  2a;?/  +  3a;^  —  83/  —  8x  =  0; 
to  classify  the  locus,  transform  and  construct  the  equation. 

{a)  To  classify.  Write  the  general  equation  and  just  below 
it  the  given  equation,  thus  : 

ay^  +  bxy  -\-  cx^  -\-  dy  -\-  ex  -}- f  =  0 

Sy""  -{-2xy  -{-Sx''  -8y  -Sx  =  0  .  .  .  (1) 

Substituting    the    co-efficients    in    the   class  characteristic 
52  _  4  ^(.^  -^g  have      5^  —  4  ac  =  4  —  36  =  —  32  ; 
hence  &^  <4  ac. 

and  the  locus  belongs  to  the  ellipse  class.  Art.  146. 

(b)  To  refer  the  locus  to  axes  such  that  the  term  containing 
xy  shall  disap)pear. 

From  Art.  138,  (5),  we  have 

b 


tan  2  ^  = 


c  —  a 


2 

hence  tan  2  6  = =  +  00, 

o  —  o 

/.  2^  =  90°  .-.  ^  =  45°  .  .  .  (2) 


EQUATION  OF   THE   SECOND  DEGREE.  179 

i.e.,  the  new  X-axis  makes  an  angle  of  -\-  45°  with  the  old 
X-axis.  Taking  now  (10),  (11),  (3),  Art.  138,  and  substitut- 
ing values,  we  have 


a'  =  ^\c  -\-  a  —  V(c  —  af  +  b~\  =  2. 

c'  =  X  ^c  +  a  -f  V(c-ay-\-b'\  =  4. 

d'  =  dcos6  —  e  sin  6  =  ^ -^2  (d  —  e)  =  0. 

e'  =  dsine  -{-ecos6  =  ^^2  (d  -\-  e)  =  —8  V2. 

Substituting    these   values   in    (6),    Art.    138,  we  have   (/ 
being  zero), 

2tf-^4.x~-8  V2T  a;  =  0  ...  (3) 

(c)    To  refer  the  locus  to  its  centre  and  axes. 

Substituting  the  values   found  above  in   (4),  Art.  139,  we 

have  ?^  = =  0. 

2  a' 

m  =  -  -^  =  8  V2  _  ^2 


Hence/"  =  a'n-  +  c'm""  +  d'n  +  e'm-\-f=  -  8,  Art.  139, 
(3). 

Substituting  this  value  of  /"  together  with  the  values  of  a' 
and  c'  found  above  in  (5),  Art.  139,  we  have 

2  y2  _|.  4  ^2  _  8  =  0, 

or  ^'  _i_  l!  =  1  .  .  .   (4) 

2^4  ^  ^ 

for  the  reduced  equation.     The  semi-axes  of  the  ellipse  are 
a  =  V2  and  &  =  2. 
(d)    To  construct. 

Draw  the  axis  OX',  making  an  angle  of  45°  with  the  old 
X-axis.  See  (b).  Draw  OY'  1  to  OX'.  The  equation  of  the 
curve  when  referred  to  these  axes  is  given  in  (3).    Constructing 


180 


PLANE  ANALYTIC  GEOMETRY. 


the  point  0'  (V2,  0)  we  have  the  centre  of  the  ellipse.  See 
(c).  Draw  O'Y"  -L  to  OX'  at  0'.  The  equation  of  the  curve 
when  referred  to  O'Y",  O'X'  as  axes  is  given  in  (4). 


Fig.  55. 


Having   the   semi-axes,  V2   and  2,  we  can  construct  the 
ellipse  by  either  of  the  methods  given  in  Art.  78. 


DISCUSSION. 

If  2/  =  0  in  (1),  we  have  for  the  X-intercepts  0,  OD, 

3 

If  cc  =  0  in  (1),  we  have  for  the  Y-intercepts  0,  OC, 

7/  ==  0,   7/  =   -  . 

If  £c  =  0  in  (3),  we  have  y  =  j-  0;  i.e.,  the  ellipse  is  tangent 
to  the  Y'-axis. 

If  y  =  0  in  (3),  we  have  for  the  X'-intercepts  0,  OB, 

X  =  0,  cc  =  2  V2. 

If  a;  =  0  in  (4),  we  have  for  the  Y"-intercepts  O'A,  O'A'. 
2/-  ±2. 


EQUATION  OF  THE  SECOND  DEGREE.  181 

If  2^  =  0  ill  (4),  we  have  for  the  X'-intercepts  O'B,  O'O, 

a;  =  -t-  V2. 
2.    Given  the  equation  t/  —  2  xy  -{-  x^  —  2  7j  —  \  =  0,  class- 
ify the  locus,  transform  and  construct  the  equation. 

(a)  To  classify. 

ay"^  +  bxy  +  cx^  -\-  dy  -\-  ex  -\-  f  =  0 
y''  —  2xy  -h  x--2y-l  =  0  ...    (1) 

hence  b"^  —  4=  ac  =  A  —  4:  =  0, 

,'.  b-  =  4:ac  ; 

hence  the  locus  belongs  to  the  parabola  class,  Art.  146. 

(b)  To  refer  the  locus  to  axes  such  that  the  term  containing 
xy  shall  disappear. 

From  Art.  138,  (5),  we  have 

tan  26  =  —^  ; 
c  —  a 

hence,  substituting 

tan  2  ^  =  — 


1  _  1  - 

.-.  ^  =  -45°  .  .  .  (2) 

Substituting  the  values  of  the  coefficients  in  (10),  (11),  (3) 

of  Art.  138,  we  have 

a'  =  ^\c  +  a  —  V(c  —  ay  +  b''\=  0. 

c'  =  :L^c^a  +  V(c  -  ay  +  b'^  ^  =  2. 

d'  =  dGosd  —  esinO  =  —2  (^  V2)  =  —  V2. 

e'  =  t^  sin  ^  +  e  cos  ^  =  —  2  (—  i  V2)     =  +  V2. 

Substituting  these  values  in   (1),  Art.  140   (since  a'  —  0), 

we  have  2x^  —  V2  y  +  V2  a;  —  1  =  0  .  .  .   (3) 

(c)    To  refer  the  p>arahola  to  a  tangent  at  the  vertex  and  the 

axis. 

Substituting  the  values  of  the  constants  in  (a),  Art.  140, 

we  have                              e'             V2  oc^  i 

m  =  —  . = =  —  .60  nearly. 

2  c'  4  ^ 

e'2  —  4  /c'  5 


182 


PLANE  ANALYTIC  GEOMETRY. 


Substituting  the  values  of  d'  and  c  in  (3),  Art.  140  (since  d) 
is  not  zero),  we  have 

a;2  =  iV2.2/  .  .  .  (4) 
for  the  reduced  equation. 
(d)    To  construct. 

V 


Fig.  56. 

Draw  OX'  making  an  angle  of  —  45°  with  the  X-axis ; 
draw  OY'  ±  to  OX'.  See  {b).  The  equation  of  the  parabola 
when  referred  to  these  axes  is  given  in  (3). 

Constructing  the  point  (—  .35,  —  .90),  we  have  the  vertex 
of  the  parabola  0'.  See  (c).  Draw  O'X"  and  O'Y"  parallel  to 
the  axes  OX',  OY'  respectively.  The  equation  of  the  parab- 
ola referred  to  these  axes  is  given  in  (4).  The  curve  can  now 
be  constructed  by  either  of  the  methods  given  in  Art.  54. 


EQUATION  OF  THE  SECOND  DEGREE.  183 

DISCUSSION. 

If  ic  =  0  in  (1),  we  have  for  the  Y-intercepts  OC,  OC, 

y  =  2.4  2/  =  -  •4- 

If  y  =  0  in  (1),  we  have  for  the  Y-intercept  OD,  OD', 

a^=  -t  1. 
If  cc  =  0  in  (3),  we  have  for  the  Y'-intercept  OK, 
1 


y  = 


=  -  .707. 


V2 
If  2/  =  0  in  (3),  we  have  for  the  X'-intercepts  OL,  OL', 

_  _  V2  +  VlO  _   -  V2  -  Vlo 

If  X  =  0  in  (4),  y  =  0;  if  ?/  =  0  in  (4),  x  =  -t  0. 

3.    Given  the  equation  y^  — 2x^  —  2'i/-\-6x  — 3  =  0,  classify 
the  locus,  transform  and  construct  the  equation. 

(a)  To  classify. 

ay'^  -\-  bxy  +  cx'^  -\-  dy  -\-  ex  +  f=  0. 
y^  — 2x^-2  y  +  6x  — 3  =  0  .  .  .  (1) 
b^-4:ac  =  8  .:  b^>4.ac; 
hence,  the  locus  belongs  to  the  hyperbola  class,  Art.  146. 

(b)  To  ascertain  the  direction  of  the  rectangular  axes  (xy 
being  wanting). 

tan  26  =  — ^—  =  -^  =  0 ; 
c  —  a        —  3 

.-.   ^  =  0; 

i.e.,  the  new  X-axis  is  parallel  to  the  old  X-axis. 

(c)  To  refer  the  hyperbola  to  its  centre  and  axes,  we  have, 
Art.  139,  (4), 


n  = 


d'  _  e' 

2^''^~~T7' 


hence  n  =  \,  m  ^=  -. 

2 

Substituting  in  the  value  of  /",  Art.  139,  (3),  we  have 
/"  =  a'n''  -f  c'm^  +  d'n  +  e'm -\- f  =  1 -^  -  2  +9-3; 

hence  f"  =  -. 

-^         2 


184 


PLANE  ANALYTIC  GEOMETRY. 


This  value,  together  with  the  values  of  a'  and  c'  in  (5),  Art. 
139,  gives  2^f-4.x^  =  -l  .  .  .  (3) 

for  the  required  equation. 

(d)    To  construct. 


Fig.  57. 

Construct  the  point  0'  (|,  1),  and  through  it  draw  O'X"  ||  to 
OX,  and  O'Y"  ||  to  OY.  The  equation  of  the  hyperbola 
referred  to  these  axes  is  given  in  (3).  AVe  see  from  this  equa- 
tion that  the  semi-transverse  axis  is  -.  Laying  off  this  dis- 
tance to  the  right  and  then  to  the  left  of  0',  we  locate  the 
vertices  of  the  curve  A,  A'. 

DISCUSSION. 

If  a;  =  0  in  (1),  we  have  for  the  Y-intercepts  OC,  OC, 

y  =  3,y  =  —1. 
If  2/  =  0  in  (1),  we  have  for  the  X-intercepts  OD,  OD', 


3  + V3 


V3 


EQUATION  OF  THE  SECOND  DEGREE.  185 

If  a;  =  0  in  (3),  we  have 

If  ?/  =  0  in  (3),  we  have  for  the  X-intercepts  O'A,  O'A', 

-^\ 

From  this  data  the  student  may  readily  determine  the 
eccentricity,  the  parameter,  and  the  focal  distances  of  the 
hyperbola. 

4.  Given  the  equation  T/^-fa;^  —  42/  +  4a;— 1  =  0,  class- 
ify the  locus,  transform  and  construct  the  equation. 

(a)  b"^  <4  ac  .-.  the  locus  belongs  to  the  ellipse  class. 

(b)  6  =  0.:  new  X-axis  is  ||  to  old  X-axis. 

(c)  (m,  n)=(-2,2)  and  /"  =  -  9 
hence  x^  -\-  y^  =  9 

is  the  transformed  equation  of  the  locus,  which  from  the  form 
of  the  equation  is  evidently  a  circle. 

(d)  Locate  the  point  (—  2,  2).  With  this  point  as  a  centre, 
and  with  3  as  a  radius,  describe  a  circle ;  it  will  be  the  re- 
quired locus. 

5.    y^  -  2  xy  -\-  x^  -  2  =  0. 

(a)  b^  ^  4:  ac  .-.  parabola  class. 

(b)  0  =  —  4:5°  .:  new  X-axis  inclined  at  an  angle  of  —  45° 
to  the  old  X-axis.     We  have  also 

a'  =  0,  c'  =  2,  d'  =  0,e'  =  0 

i.e.,  X  =  1  and  x  =  —  1  .  .  .  (1) 

are  the  equations  of  the  locus  when  referred  to  the  new  axes. 

(c)  The  construction  gives  the  lines  OX',  OY'  as  the  new 
axes  of  reference. 

Equations  (1)  are  the  equations  of  the  two  lines  CM,  CM' 
drawn  ||  to  the  Y'-axis  and  at  a  unit's  distance  from  it. 


186 


PLANE  ANALYTIC   GEOMETRY. 


Fig.  58. 


We  may  construct  the  locus  of  the  given  equation  without 
going  through  the  various  steps  required  by  the  general 
method.     Factoring  the  given  equation,  we  have 


hence 


{y  -x  +  V2)  {y  -X-  V2)  =  0 
y  =  X  —  V2  and  y  =  x  -\-  V2 


are  the  equations  of  the  locus.  Constructing  these  lines 
(OY,  OX  being  the  axes  of  reference),  we  get  the  two 
parallel  lines  CM,  CM'. 

Classify,  transform,  and  construct  each   of   the   following 
equations  : 

6.   2/'  -  2  cKy  +  a;'  +  2  2/  -  2  a;  +  1  =  0. 


7.  2/'  +  2  x?/  +  a;2  -  1  =  0. 

8.  b  y^  +  2  xy  +  5  x"  —  12  X  —  12  y 


x  =  ^^2 


2    "^   3 


EQUATION  OF  THE  SECOND  DEGREE.  187 

9.  2  1/  -\-  2  x^  —  4.y  —  4.x  -\-  1  =  0. 

^'  +  y'  =  |. 

10.  y^  +  x^  —  2  X  -\- 1  =  ().  

y  =  x^ -!,{{),()). 

11.  ?/2  _|_  ^2  _j_  2  a;  +  2  =  0. 

Imaginary  ellipse. 

12.  2/''  -  2  xy  +  a;2  -  8  a;  +  16  =  0. 

Parabola. 

13.  y^-  —  2xy  -\-x'^  —  y  -\-2x  —  l=0. 

Parabola. 

14.  4  a;y  —  2  ic  +  2  =  0. 

Hyperbola. 

15.  y~  —  2x'  -\-2y  +  !=(). 

Two  intersecting  lines. 

16.  ?/2  _  a;2  4_  2  2/  +  2  cc  -  4  =  0. 

Equilateral  hyperbola. 

17.  2/^  —  2  XT/  +  a;2  +  2  2/  +  1  =  0. 

18.  2/^  +  4  xy  +  4  a;'  —  4  =  0. 

19.  y""  —  2xy  -\-2x''  —  2y  -\-2x  =  0. 

20.  2/2  —  4  a;?/  +  4  a;2  =  0. 

21.  2/^  —  2  cc?/  —  cc2  _^  2  =  0. 

22.  2/'-ic'  =  0. 


188  PLANE  ANALYTIC  GEOMETRY. 


CHAPTER  X. 
HIGHER  PLANE    CURVES. 

147.  Loci  lying  in  a  single  plain  and  represented  by  equa- 
tions other  than  those  of  the  first  and  second  degrees  are 
called  Higher  Plane  Curves.  We  shall  confine  our  atten- 
tion in  this  chapter  to  the  consideration  of  a  few  of  those 
curves  which  have  become  celebrated  by  reason  of  the  labor 
expended  upon  them  by  the  ancient  mathematicians,  or  which 
have  become  important  by  reason  of  their  practical  value  in 
the  arts  and  sciences. 

EQUATIONS    OF   THE   THIRD   DEGREE. 

148.  The  Semi-cubic  Parabola. 

This  curve  is  the  locus  generated  by  the  intersection  of  the 
ordinate  TT^  of  the  common  parabola  with  the  perpendicular 
OP  let  fall  from  its  vertex  upon  the  tangent  drawn  at  T'  as 
the  point  of  tangency  moves  around  the  curve. 

1.    To  deduce  the  rectangular  equation. 

Let  T'  {x",  y")  be  the  point  of  tangency,  and  let  P  (cc ,  3/ ) 
be  a  point  of  the  curve. 

Let  y"^  =  4:  px  be  the  equation  of  the  common  parabola. 

Since  the  equation  of  the  tangent  line  T'M  to  the  parabola 
is  Art.  57,  (6), 

yy"  =  2p(x  +  x"), 

the   equation  of   the  perpendicular  (OM)   let  fall  from  the 
vertex  is 

y  =  —  ^  x  .  .  .  (V) 


HIGHER  PLANE   CURVES. 


189 


Fig.  59. 

Since  TT'  is  parallel  to  OY,  we  have  for  its  equation 

x=x"  .  .  .  (2) 
Combining  (1)  and  (2),  we  have 


y  = 


-  _  y 


'p 


But 

hence 


y"  =  V4:2)x"  ; 


V4  px"    „ 

y  = — ^ •  X  . 

2  p 

Squaring  and  dropping  accents,  we  have 

2/^  =  ^  .  .  .  (3) 
P 
for  the  equation  of  the  semi-cubic  parabola. 

This  curve  is  remarkable  as  being  the  first  curve  which  was 
rectified,  that  is,  the  length  of  a  portion  of  it  was  shown  to 


190 


PLANE  ANALYTIC    GEOMETRY. 


be  equal  to  a  certain  number  of  rectilinear  units.     It  derives 
its  name  from  the  fact  that  its  equation  (3)  may  be  written 

x^  —  p'^  y. 
2.   To  deduce  the  polar  equation. 

Making  x  =  r  cos  6  and  y  ^=  r  sin  Q  in  (3),  we  have,  after 
reduction, 

r  =  p  tan  ^  6  sec  ^  .  .  .  (4) 

for  the  polar  equation  of  the  curve. 

ScHOL.    Solving  (3)  with  respect  to  y,  we  have 


y 


=  Wf 


An  inspection  of  this  value  shows 

(a)  That  the  curve  is  symmetrical  with  respect  to  the 
X-axis ; 

(b)  That  the  curve  extends  infinitely  from  the  Y-axis  in 
the  direction  of  the  positive  abscissas. 

149.    To  dupjlicate  the  cube  hy  the  aid  of  the  parabola. 

Let  a  be  the  edge  of  the  given  cube.  We  wish  to  con- 
struct the  edge  of  a  cube  such  that  the  cube  constructed  on  it 
shall  be  double  the  volume  of  the  given  cube ;  i.e.,  that  the 
condition  x^  =  2  a^  shall  be  satisfied. 


Fig.  60. 


HIGHER  PLANE   CURVES. 


191 


Construct  the  parabola  whose  equation  is 

2/2  =  2  ace  ...  (1) 
Let  MPO  be  the  curve.     Construct  also  the  parabola  whose 

equation  is 

x^  =  ay  .  .  .  (2) 

Let  NPO  be  this  curve. 

Then  OA  (=  x),  the  abscissa  of  their  point  of  intersection 
is  the  required  edge.     For  eliminating  y  between  (1)  and  (2), 

we  have 

x^  =  2  a\ 

This  problem  attained  to  great  celebrity  among  the  ancient 
geometricians.  We  shall  point  out  as  we  proceed  one  of  the 
methods  employed  by  them  in  solving  it. 

150.    The  Cissoid. 

The  cissoid  is  the  locus  generated  by  the  intersection  (P)  of 
the  chord  (OM')  of  the  circle   (OMM'T)  with  the   ordinate 


Fig.  61. 


192  PLANE  ANALYTIC   GEOMETRY. 

MIST  (equal  to  the  ordinate  M'N'  let  fall  from  the  point 
M'  on  the  diameter  through  0)  as  the  chord  revolves  about 
the  origin  0. 

It  may  also  be  defined  as  the  locus  generated  by  the  inter- 
section of  a  tangent  to  the  parabola  ^/^  =  —  8  aa;  with  the 
perpendicular  let  fall  on  it  from  the  origin  as  the  point  of 
tangency  moves  around  the  curve. 

1.   To  deduce  the  rectangular  equation. 

Fh^st  Metliod.  —  Let  OT  =  2  a,  and  let  P  {x,  y)  be  any  point 
of  the  curve.  From  the  method  of  generation  in  this  case 
MN  =  WW  .-.  ON  =  N'T.  From  the  similar  triangles  ONP, 
ON'M',  we  have 

NP  :  ON  :  :  M'N' :  ON'. 


But  NP  =  y,  ON  =  X,  M'N'  =  VON' .  N'T  =  V(2  a  -  cc)  x, 
ON'  =  2a  -x\ 


.-.  y  :  X  ::  V(2  a  —  x)  x  -.2  a  —  x. 

Hence  y""  =  -~^ •••(!) 

2  a  —  X 

is  the  required  equation. 

Second  Method.  —  The  equation  of  the  tangent  line  to  the 
parabola  3/^  =  —  8  ax  is  Art.  65,  (2) 

,   2a 
y  =  —  sx  -\- 

The  equation  of  a  line  passing  through  the  origin  and  per- 
pendicular to  this  line  is 

X 

2/  =  -• 
s 

Combining  these  equations  so  as  to  eliminate  s,  we  have 

x^ 


y  = 


2  a  —  X 


for  the  equation  of  the  locus. 

This  curve  was  invented  by  Diodes,  a  Greek  mathematician 
of  the  second  century,  B.C.,  and  called  by  him  the  cissoid  from 


HIGHER  PLANE   CURVES.  193 

a  Greek  word  meaning  '*  ivy."  It  was  employed  by  him  in 
solving  the  celebrated  problem  of  inserting  two  mean  propor- 
tionals between  given  extremes,  of  which  the  duplication  of 
the  cube  is  a  particular  case. 

2.   To  deduce  the  polar  equation. 

From  the  figure  (OP,  PON)  =  (r,  0) 
we  have  also  r  =  OP  =  M'K  =  OK  -  OM'. 

But  OK  =  2  a  sec  ^  and  OM'  =  2  «.  cos  ^ ;  hence 
r  =  2  (z  (sec  6  —  cos  6), 
or  r  =  2  a  tan  6  sin  6 

is  the  polar  equation  of  the  curve. 

ScHOL.    Solving  (1)  with  respect  to  y,  we  have 


V^ 


2/=± 

An  inspection  of  this  value  shows 

(a)  That  the  cissoid  is  symmetrical  with  respect  to  the 
X-axis. 

(b)  That  X  =  0  and  x  =2  a  are  the  equations  of  its  limits. 

(c)  That  ic  =  2  a  is  the  equation  of  a  rectilinear  asymp- 
tote (SS'). 

151.    To  duplicate  the  cube  by  the  aid  of  the  cissoid. 

Let  OL,  Fig.  61,  be  the  edge  of  the  cube  which  we  wish  to 
duplicate.  Construct  the  arc  BO  of  the  cissoid,  CO  =  a 
being  the  radius  of  the  base  circle.  Lay  off  CD  =  2  CA  = 
2  a  and  draw  DT  intersecting  the  cissoid  in  B ;  draw  BO 
and  at  L  erect  the  perpendicular  LE,  intersecting  BO  in  R. 
Then  LR  is  the  edge  of  the  required  cube ;  for  the  equation 
of  the  cissoid  gives 

x^ 


y  = 


2  a  —  X 


OPTS 
hence         HB^  =  ^    (since    HB  =y,OB.=  x,  and   HT  = 
Hi 

2  a  —  x). 

The  similar  triangles  CDT  and  HBT  give 

CD  :  CT  ::  HB  :  HT. 


194 


PLANE  ANALYTIC  GEOMETRY. 


But   CD  =  2  CT   by  construction ;    hence   HB  =  2  HT 

2 
This  value  of  HT  in  the  value  of  HB^  above  gives 
2  0H3 


HB2  = 


HB 


hence  HB^  =  2  OH". 


The  triangles  OHB  and  OLR  are  similar ;  hence 

HB  :  OH  ::  LR  :  OL 

.-.  HB3:OH3::LR=^::OL3 
But  HB^  =  2  OH^,  hence  LR^  =  2  OL^ ;    whence  the  con- 
struction. 

152.    The  Witch. 


Fig.  62. 


V  HIGHER  PLANE   CURVES.  195 

The  witch  is  the  locus  of  a  point  P  on  the  produced  ordi- 
nate DP  of  a  circle,  so  that  the  produced  ordinate  DP  is  to 
the  diameter  of  the  circle  OA  as  the  ordinate  DM  is  to  the 
outer  segment  DA  of  the  diameter. 

It  may  also  be  defined  as  the  locus  of  a  point  P  on  the 
linear  sine  DM  of  an  angle  at  a  distance  from  its  foot  D  equal 
to  twice  the  linear  tangent  of  one-half  the  angle. 

1.    To  deduce  the  rectangular  equation. 

First  Method.  —  From  the  mode  of  generation,  we  have 

DP  :  OA  ::  DM  :  DA 
But  DP  =  2/.  OA  =  2  a,  DM  =  VOD  .  DA  =  ^/x  (2  a  -  x), 

DA  =  2  a  —  x\ 


hence  '     y-2  a  ::  V(2  a  —  x)  x  :  2  a  —  x. 

o  ^  Cv  JO  /^  \ 

2  a  —  X 
is  the  required  equation. 

Second  Method.  —  Let  MCO  =  6 ;  then  by  definition 

o     ^        ^       c,     ^  la  0-  —  cos  &) 
V  =  2  «  tan  -  =  2  ft  i  /  — ^^ '-  . 

2  V  a  (1  +  cos  6) 

But  ft  (1  —  cos  ^)  =  a,  -  a  cos  ^  =  00  —  DC  =  OD  =  x,  and 
a  (1  +  cos  ^)  =  a  +  ft  cos  ^  =  00  +  DC  =  OD'  =  2a  —  x; 


hence 

4  a^x 


=  2^2- 


or,  squaring        y 


2  a  —  X 

This  curve  was  invented  by  Donna  Maria  Agnesi,  an  Italian 
mathematician  of  the  eighteenth  century. 

ScHOL.    Solving  (1)  with  respect  to  y,  we  have 


y 


=  _1_  2  a  \/ ^- . 

\    2a-x 


Hence    (a)    the  witch  is  symmetrical  with  respect  to  the 
X-axis. 

(b)  cc  =  0  and  x  =  2  a  are  the  equations  of  its  limits. 

(c)  x  =  2a  is  the  equation  of  the  rectilinear  asymptote  SS'. 


196 


PLANE  ANALYTIC  GEOMETRY, 


EQUATIONS   OF  THE   FOURTH   DEGREE. 

153.    The  Conchoid. 

The  conchoid  is  the  locus  generated  by  the  intersection  of 
a  circle  with  a  secant  line  passing  through  its  centre  and  a 
fixed  point  A  as  the  centre  of  the  circle  moves  along  a  fixed 
line  OX. 

As  the  intersection  of  the  circle  and  secant  will  give  two 
points  P,  P,  one  above  and  the  other  below  the  fixed  line,  it 
is  evident  that  during  the  motion  of  the  circle  these  points 
will  generate  a  curve  with  two  branches.  The  upper  branch 
MBM'  is  called  the  Supekior  Branch  ;  the  lower,  the  In- 
ferior Branch.  The  radius  of  the  moving  circle  O'P 
(=  OB)  is  called  the  Modulus.  The  fixed  line  OX  is  called 
the  Directrix  ;  the  point  A,  the  Pole. 


1,   To  deduce  the  rectangidar  equation. 

Let  P  (x,  y),  the  intersection  of  the  circle  PP'P  and  the 


HIGHER  PLANE   CURVES.  197 

secant  AOT,  be  any  point  of  the  curve.  Let  O'P  =  OB  =  h, 
and  let  OA  =  a. 

The  equation  of  the  circle  whose  centre  is  at  0'  (x',  0)  is 

(x  —  x'Y  -\-  y^  =  b^- 
The  equation  of  the  line  AOT  is 
y  ^=  sx  —  a  .  .  .   (1) 
Making  y  —  0  in  (1),  we  have 

a 

s 
for  the  distance  00'. 
But  00'  =  x' ;  hence 

^-^J+^'  =  ^'-  •  •   ^^> 

is  the  equation  of  the  circle.  If  we  now  combine  (1)  and  (2) 
so  as  to  eliminate  s,  the  resulting  equation  will  express  the 
relationship  between  the  co-ordinates  of  the  locus  generated 
by  the  intersection  of  the  loci  they  represent.  Substituting 
the  value  of  s  drawn  from  (1)  in  (2),  we  have 


a  +y 

,.xY=(b'-y^)(a  +  yy  .   .  .   (3) 
is  the  required  equation. 

We  might  have  deduced  this  equation  in  the  following  very 
simple  way :  Draw  AT  ||  to  OX,  and  PT  ||  to  OY.     Since  the 
triangles  ATP  and  O'SP  are  similar,  we  have 
PS  :  SO'  ::  PT  :   TA  : 


i.e.,  y  :  -\/b^  —  y^  ::  a  -{-  y  :  x. 

Hence  xY  =  {b^  —  y^)  (a  +  yy. 

This  curve  was  invented  by  Nicomedes,  a  Greek  mathema- 
tician who  flourished  in  the  second  century  of  our  era. 

It  was  employed  by  him  in  solving  the  problems  of  the 
duplication  of  a  cube  and  the  trisection  of  an  angle. 


198  PLANE  ANALYTIC  GEOMETRY. 

2.   To  deduce  the  polar  equation. 

From  the  figure  we  have  (AY  being  the  initial  line,  and  A 
the  pole) 

(AP,  PAB)  =  (r,  6) 
But  AP  =  AO'  ±  OT ; 

hence  r  =  a  sec  6  ^^b 

is  the  polar  equation  of  the  curve. 

ScHOL.    Solving  (3)  with  respect  to  x,  we  have 

O'  +  y 


X  =  ^  —       ■y/b''  —  if  .     ■ 

An  inspection  of  this  value  shows 

(a)  That  the  conchoid  is  symmetrical  with  respect  to  the 
Y-axis. 

(5)    That  y  ^b  and  y  =  —  b  are  the  equations  of  its  limits. 

(c)  That  2/  =  0  gives  x  =  -I-  c/d,  .-.  the  X-axis  is  an  asymp- 
tote. 

(d)  If  a  =  0,  then  x  =  -^  V^"  —  y'';  i.e.,  the  conchoid  be- 
comes a  circle. 

(e)  If  6  >  a,  the  inferior  branch  has  a  loop  as  in  the  figure, 
(/)    li  b  =  a,  the  points  A'  and  A  coincide  and  the  loop. 

disappears. 

(g)  li  b  <C  a,  the  inferior  branch  is  similar  in  form  to  the 
superior  branch,  and  the  point  A  (o,  —  a)  is  isolated ;  i.e., 
though  entirely  separated  from  the  curve,  its  co-ordinates  still 
satisfy  the  equation. 

154.    To  trisect  an  angle  by  the  aid  of  the  conchoid. 

Let  PCX  be  the  angle  which  we  wish  to  trisect.  Prom  C 
with  any  radius  as  CD  describe  the  semi-circle  DAH.  From 
the  point  A  draw  AB  1  to  CX  and  make  OB  =  CD.  With 
A  as  a  pole  and  OB  as  a  modulus  construct  a  conchoid  on 
CX  as  a  directrix.  Join  H,  the  intersection  of  the  inferior 
branch  and  the  circle,  with  A  and  produce  it  to  meet  the 
directrix  in  K  ;  then 

CKA  =  -  PCX. 
3 


HIGHER  PLANE   CURVES. 


199 


Fig.  64. 


For  join  H  and  C ;  then  from  the  nature  of  the  conchoid 
HK  =  HC  =  OB. 
.  From  the  figure  PCX  =  CAK  +  CKA ; 
but  CAK  =  CH A  =  2  CKA ; 

hence  PCX  =  2  CKA  +  CKA. 

Therefore       CKA  =  ^  PCX. 

We  might  have  used  the  superior  branch  for  the  same  pur- 
pose. 

155.    The   Limacon. 
Y 


Fig.  65. 


200  PLANE  ANALYTIC  GEOMETRY. 

The  lima9on  is  the  locus  generated  by  the  intersection  of 
two  lines  OP,  CP  which  are  so  related  that  during  their  revo- 
lution about  the  points  0  and  C  the  angle  PCX  is  always 
equal  to  |  POX. 

1.    To  deduce  the  polar  equation. 

Let  0  be  the  pole,  and  OX  the  initial  line.     Let  P  be  any 
point  of  the  curve,  and  let  OC  =  a ;  then 
(OP,  POX)  =  (r,  B). 

From  the  triangle  POC,  we  have 

OP:OC::sin  OOP:  sin  OPC  ; 

i.e.,  v.  a::  sin  |  B  :  sin  ^  $. 

TT  (f'  sin  3-  $ 

Hence  r  = ^— 

sin  i-  ^    ■ 

From  Trigonometry 

sin  I  ^  =  3  sin  ^  ^  -  4  sin^  |  ^  =  (3  -  4  sin^  ^0)sin^e; 

hence  r  =  a.  (3  —  4  sin^  ^  0), 

_.Y3-4l-^°« 


2 

=  a  (1  +  2  cos  6)  .  .  .  (1) 
is  the  polar  equation  of  the  lima9on. 
2.    To  deduce  the  rectangular  equation. 
From  Art.  35,  we  have 


=  Va^^  +  y",  cos  6  = 


for  the  equations  of  transformation  from  polar  to  rectangular 
co-ordinates.     Substituting  these  values  in  (1),  we  have 


Vic"^  +  y^ 


-y/x^  -(-  y~ 

or  (x^  +  2/^-2  axy  =  a'  (x^  +  y'-)  ...   (2) 

for  the  required  equation. 

ScHOL.  1.    From  the  triangle  ODA,  we  have 

OD  =  OA  cos  e  =  2a  cos  0. 
From  (1)  OP  =  a  +  2  a  cos  0 ; 

hence  OP  -  OD  =  DP  =  a ; 


HIGHER  PLANE   CURVES. 


201 


i.e.,  the  intercept  between  the  circle  ODA  and  the  lima9on 
of  the  secant  through  0  is  constant  and  equal  to  the  radius  of 
the  circle. 

ScHOL.  2.        If  ^  =  0,  r  =  3  a  =  OB. 

If  ^  =  90°,  r  =  a  =  OM. 

If  ^  =  180°,  r  =  -  a  =  OG 

If  ^  =  270°,  r  =  a  =  OM' 

156.    The  Lemniscata. 

The  lemniscata  is  the  locus  generated  by  the  intersection  of 
a  tangent  line  to  the  equilateral  hyperbola  with  a  perpen- 
dicular let  fall  on  it  from  the  origin  as  the  point  of  tangency 
moves  around  the  curve. 


Fig.  66. 

1.    To  deduce  the  rectangular  equation. 

Since  T  {x" ,  y")  is  a  point  of  the  equilateral  hyperbola,  we 
have,  Art.  103,  Cor  1, 

x"^  -  y"^  ==a''  .  .  .   (1) 

The  equation  of  the  tangent  line  TP  is,  Art.  112, 
XX"  -  yy"  =  «^  .  .  .   (2) 


202  PLANE  ANALYTIC  GEOMETRY. 

x" 
Since  the  slope  of  this  line  is  — ,  the  equation  of  the  per- 

y 

pendicular  OP  is 

v" 

2/  =  _  ^  a;  .  .  .   (3) 

X 

Treating  (2)  and  (3)  as  simultaneous  and  solving  for  x"  and 
y",  we  find 


and  y"  =  — 


OX 


^2  _j_  y2  "  a;2  _|_  y2 

Substituting  these  values  in  (1),  we  have 

4      9  4      9 

a^x'^  —  ay-         » 
(x'  +  ff 
or.  {x^  +  t/y  =  a^  (x"  -if)  .  .  .  (4) 

for  the  required  equation. 

This  curve  was  invented  by  James  Bernouilli.  It  is  quad- 
rable,  its  area  being  equal  to  the  square  constructed  on  the 
semi-transverse  axis  OA. 

2.    To  deduce  the  polar  equation. 

We  have  Art.  34,  (3),  for  the  equations  of  transformation 
X  =  r  cos  6,  y  =  r  sin  9. 

These  values  in  (4)  give 

\ ?-2  (cos^  $  +  sin^  6) l^=a^\ r^  (cos^  0  -  sin^  6)} ; 
therefore  r*  =  a^  r"^  cos  2  6, 

or   '  r^  =  0,2  gQg  2  6...    (5) 

is  the  required  equation. 

ScHOL.    If  ^  =  0,  cos  20  =  cos  0  =  1 .  •.  r  =  -1^  a. 

If  ^  <  45°,  cos  2  6  <.  cos  90°  .-.  r  has  two  equal  values  with 
opposite  signs. 

If  ^  =  45°,  cos  ?  ^  =  cos  90°  =  0  .-.  r  =  0. 

If  ^  >  45°  and  <135°  r  is  imaginary. 

If  ^  =  135°,  cos  2  ^  =  cos  270°  =  0  .-.  r  =  0. 

If  ^  =  180°,  cos  2  ^  =  cos  360°  =  1  .-.  r  =  -J-  a. 

An  examination  of  these  values  of  r  shows  that  the  curve 
occupies  the  opposite  angles  formed  by  the  asymptotes  of  the 
hyperbola. 

The  curve  is  symmetrical  with  respect  to  both  axes. 


HIGHER  PLANE   CURVES. 


203 


TRANSCENDENTAL   EQUATIONS. 

157.    The  Curve  of  Sines. 
This  curve  takes  its  name  from  its  equation 
y  =  sin  X, 
and  may  be  defined  as  a  curve  whose  ordinates  are  the  sines 
of  the  corresponding  abscissas,  the  latter  being  considered  as 
rectified  arcs  of  a  circle. 
Y 


Fig.  67. 

To  construct  the  curve.  Give  values  to  x  which  differ  from 
each  other  by  30°,  and  find  from  a  "Table  or  Natural 
Sines  "  the  values  of  the  corresponding  ordinates. 

Tabulating  the  result,  we  have, 

Value  of  X  Corresponding  Value  of  y 

0 

.50 


.87 


30° 

= 

0 

IT 

6  ~ 

.52                    « 

60° 

= 

2  TT 

"6" 

=  1.04               « 

90° 

= 

37r 

6 

=  1.56                « 

20° 

47r 

6 

=  2.08 

1.00 


.87 


204 


PLANE  ANALYTIC  GEOMETRY. 


Value  of  X 

Corr 

esponding 

Value  of  y 

150°  -^"^  -  2.60 
6 

« 

.50 

180°  =  TT  =  3.14 

(( 

0 

210°  =  IjL  =  3.66 
6 

(( 

-.50 

240°  =  ?Z  =  4.18 
6 

It 

-.87 

270°  -  ^  ^  -  4.70 
6 

u 

-  1.00 

300°  =  ^^  '^  -  5.22 
6 

11 

-.87 

330°  =  ^^'^  =  5.75 
6 

a 

-.50 

360°  =  2  TT  =  6.28 

a 

0 

Constructing  these  points  and  tracing  a  smooth  curve 
through  them,  we  have  the  required  locus.  As  x  may  have 
any  value  from  0  to  -J:;  oo  and  yet  satisfy  the  equation  of  the 
curve,  it  follows  that  the  curve  itself  extends  infinitely  in 
the  direction  of  both  the  positive  and  negative  abscissas. 

158.   The  Cukve  of  Tangents. 
Y 


Fig. 


HIGHER  PLANE   CURVES.  205 

This  curve  also  takes  its  name  from  its  equation 
y  =  tan  x. 

To  construct  the  curve.  Give  x  values  differing  from  each 
other  by  30°  and  find  from  a  Table  of  Natural  Tangents  the 
corresponding  values  of  y.     Tabulating,  we  have, 

Value  of  X  Corresponding  Value  of  y 

0  "  0 

"  .57 

«  1.73 


30° 

"  6  ~ 

--   .52 

60° 

_27r 

6 

=  1.04 

90° 

377 

6 

=  1.56 

120° 

_47r 

"6" 

=  2^8 

150° 

_57r 

6 

=  2.60 

180° 

=  TT  = 

3.14 

210° 

_lir 

=  3.66 

«  -  1.73 

-.57 

"  0 

.57 
6 

240°  =^  =  4.18  "  1.73 

6 

270°  =  ^  =  4.70  «  oo 

6 

300°  =  ^^  =  5.22  "  -  1.73 

6 

330°  =  11^  ==  5.75  "  -  .57 

6 

360°  =  2  TT  =  6.28  "  0 

Constructing  these  points  and  tracing  a  smooth  curve 
through  them,  we  have  the  locus  of  the  equation. 

This  curve,  together  with  that  of  the  preceding  article, 
belong  to  the  class  of  Repeating  Curves,  so  called  because 
they  repeat  themselves  infinitely  along  the  X-axis. 


206 


PLANE  ANALYTIC  GEOMETRY. 


159.    The  Cycloid. 

This  curve  is  the  locus  generated  by  a  point  on  the  circum- 
ference of  a  circle  as  the  circle  rolls  along  a  straight  line. 
The  line  OM  is  called  the  Base  of  the  cycloid ;  the 


point  P,  the  Generating  Point  ;  the  circle  BPL,  the  Gen- 
erating Circle;  the  line  HB',  perpendicular  to  OM  at  its 
middle  point,  the  Axis.  The  points  0  and  M  are  the  Vertices 
of  the  cycloid. 

1.  To  deduce  the  rectangular  equation,  the  origin  being 
taken  at  the  left-hand  vertex  of  the  curve. 

Let  P  be  any  point  on  the  curve,  and  the  angle  through 
which  the  circle  has  rolled,  PCB  =  0.  Let  LB,  the  diameter 
of  the  circle,  =  2  a. 

Then  OA  =  OB  -  AB  and  AP  =  CB  -  CK. 

ButOA  ==  a;,  OB  =  a  ^,  AB  =  PK  =  a  sin  ^,  AP  =  3/,  CB  =  a, 

CK  =  a  cos  9 ;  hence,  substituting,  we  have 


X  =  a  t)  —  a  sm 
y  =z  a  —  a  cos  6 


(1) 


HIGHER  PLANE   CURVES.  207 


Eliminating  6  between  these  equations,  we  have 

/T     If  1i 

X  =  a  cos"^  —  V'2  ay  —  y"^  =  a  vers~^  - 

CO  (^ 


-^2ay-y'  ...  (2) 
for  the  required,  equation. 

ScHOL.    An  inspection  of  (2)  shows 

{a)   that  negative  values  of  y  render  x  imaginary. 

(b)  When  y  =  0,  x  =  a  vers~^  0  =  0;  but  a  vers"^  0  =  2  tt  a, 
or  4  TT  a,  or  6  TT  (X,  or  etc. ;  hence  there  are  an  infinite  number 
of  points  such  as  0  and  M. 

(c)  When  y  =  2  a,  x  =  a  vers"^  2  =  tt  a  =  OB' ;  but 
a  vers~^  2  =  3  tt  a,  or  5  tt  a,  ov  1  tt  a,  ov  etc. ;  hence,  there  are 
an  infinite  number  of  points  such  as  H. 

(d)  y  =  0  and  y  =  2  a  are  equations  of  the  limits. 

(e)  For  every  value  of  y  between  the  limits  0  and  2  a  there 
are  an  infinite  number  of  values  for  x. 

2.  To  deduce  the  rectangular  equation,  the  origin  being  at  the 
highest  'point  H. 

We  have  for  the  equations  of  transformation 

a;  =  OA  =  OB'  -  PK'  =  tt  a  +  cc' 

2/  =  AP  =  B'H  -  HK'  =  2a^y' 
These  values  in  (1)  above  give 

x'  =^  a  {0  —  it)  —  a  sin  Q  1  .o\ 

y'  '=  —  a  —  a  cos  Q  f 

But  &,  the  angle  through  which  the  circle  has  rolled  from 
H,  =  ^  —  TT  ;  hence 

x'  =  aQ'  -\-  a  sin  &  1  ,^ 

y'  =  a  (cos  6'  -1)  ]    ■  ■  '   ^  ^ 

_   y'  _ . 

Hence  cc'  =  a  vers~^  — ■  -}-  V—  2  ay^—  y'^ .  .  .  (5) 

The  invention  of  this  curve  is  usually  attributed  to  Galileo. 
With  the  exception  of  the  conic  sections  no  known  curve 
possesses  so  many  useful  and  beautiful  properties.  The  fol- 
lowing are  some  of  the  more  important : 


208 


PLANE  ANALYTIC  GEOMETRY. 


1.  Area  OPHDB'O  =  area  HDB'  =  tt  a". 

2.  Area  of  cycloid  OHMO  =  3  HDB'  =  3  tt  a^. 

3.  Perimeter  OPHM  =  4  HB'  =  8  a. 

4.  If  two  bodies  start  from  any  two  points  of  the  curve 
(the  curve  being  inverted  and  friction  neglected),  they  will 
reach  the  lowest  point  H  at  the  same  time. 

5.  A  body  rolling  down  this  curve  will  reach  the  lowest 
point  H  in  a  shorter  time  than  if  it  were  to  pursue  any  other 
path  whatever. 

SPIRALS. 

160.  The  Spiral  is  a  transcendental  curve  generated  by  a 
point  revolving  about  some  fixed  point,  and  receding  from  it 
in  obedience  to  some  fixed  law. 

The  portion  of  the  locus  generated  during  one  revolution  of 
the  point  is  called  a  Spire. 

The  circle  whose  radius  is  equal  to  the  radius-vector  of  the 
generating  point  at  the  end  of  the  first  revolution  is  called 
the  Measuring  Circle  of  the  spiral. 

161.  The  Spiral  of  Archimedes. 


\^^— 

r-x  / 

/\    "''' ' 

~~  ""■•^w\ 

/    ''   \ 

x\\ 

/      /      \^-- 

/       \ 

/        1        /     \ 

y         'a 

\  [/ 

\    // 

\                        ^^          J/  \ 

\  /       * 

\                    .X           ^^''^>-^ 

^^^-^""^^ 

\/    '^-— 

■"    \ 

Fig.  70. 


HIGHER  PLANE   CURVES.  209 

This  spiral  is  the  locus  generated  by  a  point  so  moving 
that  the  ratio  of  its  radius-vector  to  its  vectorial  angle  is 
always  constant. 

From  the  definition,  we  have 


hence  r  ■=  cO  .  .  .   (1) 

is  the  equation  of  the  spiral. 

To  construct  the  spiral. 

Assuming  values  for  6  and  finding  from  (1)  the  correspond- 
ing value  for  r,  we  have 

Values  of  6  Corresponding  Values  of  r 

0  "0 

"  .  —  C 

4 

cc  2  TT 

4 

,,  3  TT 

"  G 

4 

"  TTC 

4 

"  c 

4 

"  . G 

4 

"  2  TTC 

Constructing  these  points  and  tracing  a  smooth  curve 
through  them,  we  have  a  portion  of  the  spiral. 

Since  ^  =  0  gives  r  =0,  the  spiral  passes  through  the  pole. 

Since  $  =  cc  gives  r  =  oo,  the  spiral  makes  an  infinite 
number  of  revolutions  about  the  pole. 

Since  0  =  2  tt  gives  r  =  2  ir  c,  OA  (=  2  tt  c)  is  the  radius  of 
the  measuring  circle. 


45° 

TT 

"4 

90° 

_27r 
4 

135° 

_37r 

4 

180° 

=   TT 

225° 

5x 
4 

270° 

_67r 

7  _ 

315° 

/    TT 

4 

360° 

=   277- 

00 

210 


PLANE  ANALYTIC  GEOMETRY. 


162.    The  Hyperbolic  Spiral. 

This  curve  is  the  locus  generated  by  a  point  so  moving  that 
the  product  of  its  radius-vector  and  vectorial  angle  is  always 
constant. 

From  the  definition  we  have 


or  r  =  -  .  .  .   (1) 

e  ^  ^ 

for  the  equation  of  the  spiral. 


Fig.  71. 
To  construct  the  spiral. 

Giving  values  to  6,  finding  the  corresponding  values  of  r, 
we  have 

Values  of  6  Corresponding  Values  of  r 


0 

45°  =  '- 


00 

n 


HIGHER  PLANP   CURVES.  211 


Values  of  6 

Corresponding 

Values  of  r 

90°    =  ?  TT 

4 

a 

2c 

TT 

135°=^7r 

4 

ti 

4c 

3  TT 

180°    =  TT 

u 

c 

IT 

225°  =  ^  TT 
4 

u 

Ac 
57r 

270°=^7r 
4 

u 

4c 

315°  =1,7 
4 

it 

4c 

360°  =  2  TT 

ii 

2^ 

00, 

a 

0 

Constructing  the  points  we  readily  find  the  locus  to  be  a 
curve  such  as  we  have  represented  in  the  figure. 

Since  ^  =  0  gives  r  =  oo  there  is  no  point  of  the  spiral 
corresponding  to  a  zero-vectorial  angle. 

Since  6  ^  oo  gives  r  =  0,  the  spiral  makes  an  infinite  number 
of  revolutions  about  the  pole  before  reaching  it. 

Since  6  =  2tt  gives 

c 

2  TT 

c  is  the  circumference  of  the  measuring  circle. 
ScHOL.    Let  P  be  any  point  on  the  spiral ;  then 

(OP,  POA)  =  (r,  0). 

With  0  as  a  centre  and  OP  as  a  radius  describe  the  arc  PA. 
By  circular  measure,  Arc  PA  =  r  6,  and  from  (1)   c  =r  0; 
hence  Arc  PA  =  c  ; 

i.e.,  the  arc  of  any  circle  between   the  initial  line  and  the 
spiral  is  equal  to  the  circumference  of  the  measuring  circle. 


212 


PLANE  ANALYTIC  GEOMETRY. 


163.    The   Parabolic    Spiral. 

This  spiral  is  the  locus  generated  by  a  point  so  moving 
that  the  ratio  of  the  square  of  its  radius-vector  to  its  vectorial 
angle  is  always  constant. 

From  the  definition  we  have 


or,  r^  =  c6  .  .  .  (1) 

for  the  equation  of  the  spiral. 


Fig.  72. 

To  construct  the  spiral. 

Values  of  $ 
0 

Corresponding 

Values  of  r 
0 

45°='" 
4 

11 

^  Vctt 

90°  _  2  T 
4 

ti 

i  V2c7r 

135°  -  ^7 
4 

ti 

•1  VSCTT 

180°   =  TT 

li 

Vc  TT 

HIGHER  PLANE   CURVES. 


213 


Values  of  $ 

225°  =  ^ 
4 

270^  =  — 
4 

315°  =  1^ 
4 

360°  =  2  TT 

GO 


Corresponding 


Values  of  r 


*V6 


Ctt 


V27^ 
00 


Constructing  these  points  and  tracing  a  smooth  curve 
through  them  we  have  the  required  locus. 

Since  ^  =  0  gives  r  =  0,  the  spiral  passes  through  the  pole. 

Since  6  =  cc  gives  r  =  go,  the  spiral  has  an  infinite  num- 
ber of  spires. 

164.    The  Lituus  or  Trumpet. 

This  curve  has  for  its  equation 


or 


=  v/,^..w 


Fig.  73. 


214 


PLANE  ANALYTIC  GEOMETRY. 


If  ^  =  0,  ?'  =  00 ;  if  ^  =  00,  r  =  0.  This  curve  has  the 
initial  line  as  an  asymptote  to  its  infinite  branch. 

165.    The  Logarithmic  iSph^al. 

This  spiral  is  the  locus  generated  by  a  point  so  moving  that 
the  ratio  of  its  vectorial  angle  to  the  logarithm  of  its  radius 
vector  is  equal  to  unity.     Hence 

=  1;  i.e.,  e  =  logr; 


log  r 


or  passing  to  equivalent  numbers  (a  being  the  base),  we  have 

r  ^aO  .  .  .  (1) 
for  the  equation  of  the  spiral. 


To  construct  the  sjnral.     Let  a  =  2,  then 
r  =  29 
is  the  particular  spiral  we  wish  to  construct. 


HIGHER  PLANE   CURVES.  215 


Values  of  ^ 

Corresponding 

Values  of  r 

0 

a 

1 

1  =  57.°3 

u 

2 

2  =  114.°6 

ii 

4 

3  =  171.°9 

a 

8 

4  _  229.  °2 

a 

16 

00 

i( 

00 

-  1  =  -  57.°3 

(I 

.5 

-  2  =  -  114. 

°6 

i( 

.25 

-  3  =  -  171. 

°9 

ii 

.125 

_.  4  =  —  229. 

°2 

(( 

.062 

—  00 

i(  ' 

0 

A  smooth  curve  traced  through  these  points  will  be  the 
required  locus. 

Since  6  =  0  gives  r  =  1  whatever  be  the  assumed  value  of 
a,  it  follows  that  all  logarithmic  spirals  must  intersect  the 
initial  line  at  a  unit's  distance  from  the  pole. 

Since  6  =  cc  gives  ?•  =  oo ,  the  spiral  makes  an  infinite 
number  of  revolutions  without  the  circle  whose  radius  OA  =  1, 

Since  6  =  —  oc  gives  r  =  0,  the  spiral  makes  an  infinite 
number  of  revolutions  within  the  circle  OA  before  reaching 
its  pole. 

EXAMPLES. 

1.  Discuss  and  construct  the  cubical  parabola 

_  ^^ 

2.  What  is  the  polar  equation  of  the  limayon,  Fig,  65,  the 
pole  being  at  C  ? 

Ans.    r  =  2  a  cos  -  Q. 
o 

3.  Let  OF  =  OF'  =  a  yi  Fig.  ^^.  Show  that  the  lemnis- 
cata  is  the  locus  generated  by  a  point  so  moving  that  the 


216  PLANE  ANALYTIC   GEOMETRY. 

product  of  its  distances   from  the  two  fixed  points  F,  F'  is 
constant  and 


Discuss  and  construct  the  loci  of  the  following  equations 

4.  a;  =  tan  y. 

5.  1/  =  cos  X. 

G.  y  =  sec  X. 
7.  a;  =  sin  y. 
S.    y  =  cot  X. 

2.   y  =  cosec  X. 

10-  y  =  ^ 

11.   xV +  :.,/  =  !.  _..    ,        j_^i„^- 

20.   Discuss  and  construct  the  locus  of  the  equation 

yi  _  96  ah/  +  100  a'^x'^  —  x*  =  0  or 

y  =   ^  Vis  «2  _^  V(a;  —  6  a)  (x  +  6  a)  (x  —  S  a)  (x -{- 8  a). 

21.  Show  that  ?/  =  _j-  cc  are  the  equations  of  the  rectilinear 
asymptotes  of  the  locus  represented  by  the  equation  of 
Ex.  20. 


12. 

a^  =  x^  —  axy. 

13. 

xl  +  ?/§  =  1. 

14. 

15. 

r  ^=  a  sin  2  6. 

16. 

a 
^  ~  sin  2  ^  ' 

17. 

r  =  a  sm**  -  . 
3 

18. 

r2  sin2  2^  =  1. 

1Q 

..  _  1  +  sin  6 

N 


SOLID   ANALYTIC   GEOMETRY. 
PART    11. 


CHAPTER  I. 
CO-ORDINATES. —THE    TRI-PLANAR    SYSTEM.. 

166.  The  position  of  a  point  in  space  is  determined  when 
we  know  its  distance  and  direction  from  three  planes  which 
intersect  each  other,  these  distances  being  measured  on 
lines  drawn  from  the  point  parallel  to  the  planes.  Although 
it  is  immaterial  in  principle  what  angle  these  planes  make 
with  each  other,  yet,  in  practice,  considerations  of  convenience 
and  simplicity  have  made  it  usual  to  take  them  at  right 
angles.     They  are  so  taken  in  what  follows. 

Let  XOZ,  ZOY,  YOX  be  the  Co-ORDiisrATE  Planes  inter- 
secting each  other  at  right  angles.  Let  OX,  OY,  OZ  be  the 
Co-OEDiNATE  AxES  and  0,  their  intersection,  the  Origin  of 
Co-ordinates. 

Let  P  be  any  point  in  the  right  triedral  angle  0  -  XYZ. 
Then  P  is  completely  determined  when  we  knoAV  the  lengths 
and  directions  of  the  three  lines  PA,  PB,  PC  let  fall  from 
this  point  on  the  planes. 

As  the  planes  form  with  each  other  eight  right  triedral 
angles,  there  are  evidently  se^wn  other  points  which  satisfy 
the  condition  of  being  at  these  distances  from  the  co-ordi- 
nate planes.     The  ambiguity  is  avoided  here  (as  in  the  case 

217 


218 


SOLID  ANALYTIC   GEOMETRY. 


of  the  point  in  a  plane)  by  considering  the  directions  in  which 
these  lines  are  measured. 

Assuming  distances  to  the  right  of  YOZ  a,s  positive,  distances 
to  the  left  will  be  negative. 


Pa 

P3 

p. 

/j 

l\ 

/ 

/\\                        B/ 

'1    ,.       P/ 

1 _ 

]/    1 

/              / 

1     /             /    ' 

1       /    V            /         ' 

!/     / 

i 

-X 

/'  1                n/ 

'T-----.  / 

+x 

i                      c 

-■) 

f 

/ 

F 

3 

-2 

p. 

Fig.   75. 


Assuming  distances  above  XOY  as  j^ositive,  distances  below 
will  be  negative. 

Assuming  distances  in  front  o/XOZ  as  J9os^^5^^' e,  distances  ifo 
^7ie  rear  will  be  negative. 

Calling  x',  y',  z'  (=  BP,  AP,  CP,  respectively)  the  co-ordi- 
nates of  the  point  P  in  the  first  angle,  we  have  the  follow- 
ing for  the  co-ordinates  of  the  corresponding  points  in  the 
other  seven : 

Second  Angle,  above  XY  plane,  to  left  YZ  plane,  in  front 
of  XZ  plane,  (-  x',  y',  z')  V^. 

Third  Angle,  above  XY  plane,  to  left  YZ  plane,  in  rear 
of  XZ  plane,  (-  x',  -  y',  z')  P3. 


CO-ORDINA  TES.  219 

Fourth  Angle,  above  XY  plane,  to  right  YZ  plane,  in 
rear  of  XZ  plane,  {x',  —  y',  z')  P4. 

Fifth  Angle,  below  XY  plane,  to  right  YZ  plane,  in 
front  of  XZ  plane,  (x',  y',  -  z')  P5. 

Sixth  Angle,  below  XY  plane,  to  left  YZ  plane,  in  front 
of  XZ  plane,  (—  x',  y',  —  z')  Pg. 

Seventh  Angle,  below  XY  plane,  to  left  YZ  plane,  in 
rear  of  XZ  plane,  {—  x',  —  y',  —  z')  P7. 

Eighth  Angle,  below  XY  plane,  to  right  YZ  plane,  in  rear 
of  XZ  plane,  {x',  —  /,  —  z')  Pg. 


EXAMPLES. 

1.  In  what  angles  are  the  following  points  : 

(1,  2,  -  3),  (-  1,  3,  -  2),  (-  1,  -  2,  -  4),  (3,  -  2,  1). 

2.  State  the  exact  position  with  reference  to  the  co-ordi- 
nate axes  (or  planes)  of  the  following  points  : 

(0,  0,  2),  (-  2,  1,  2),  (3,  1,  0),  (3,  -  1,  2),  (2,  0,  3),  (-  1,  2, 
0),  (0,  -  1,  0),  (3,  0,  1),  (1,  -  2,  3),  (0,  0,  -  2),  (4,  1,  2), 
(5,  1,  -  1),  (1,  1,  -  1). 

3.  In  which  of  the  angles  are  the  X-co-ordinates  positive  ? 
In  which  negative  ?  In  which  of  the  angles  are  the  Y-co- 
ordinates  positive  ?     In  which  are  the  Z-co-ordinates  negative? 

167.  Projections.  The  projection  of  a  point  on  a  plane  is 
the  foot  of  the  perpendicular  let  fall  from  the  point  on  the 
plane.  Thus  A,  B,  and  C,  Fig.  75,  are  the  projections  of  the 
point  P  on  the  planes  XZ,  YZ,  XY,  respectively. 

The  projection  of  a  line  of  definite  length  on  a  plane  is  the 
line  joining  the  projections  of  its  extremities  on  that  plane. 
Thus  OC,  Fig.  75,  is  the  projection  of  OP  on  the  XY  plane. 

The  projection  of  a  line  of  definite  length  on  another  line 
is  that  portion  of  the  second  line  included  between  the  feet  of 
the  perpendiculars  drawn  from  the  extremities  of  the  line  of 
definite  length  to  that  line. 


220 


SOLID  ANALYTIC  GEOMETRY. 


Thus  OM,  Fig.  75,  is  the  projection  of  OP  on  the  X-axis. 

KoTE.  —  The  projections  of  points  and  lines  as  above  de- 
fined are  orthogonal.  Unless  otherwise  stated,  all  projections 
will  be  so  understood  in  what  is  to  follow. 

168.    To  find  the  length  of  a  line  joinmff  two  points  in  space. 


Fig.  76. 

Let  P'  {x',  y',  z')  and  Y'  {x",  y",  z")  be  the  given  points. 

Let  L  (=  PT")  be  the  required  length.  Draw  Y'G  and 
P'N  II  to  OZ;  NA  and  CD  ||  to  OY;  NB  ||  to  OX.  Join  N 
and  C  and  draw  P'M  ||  to  NC. 

We  observe  from  the  figure  that  L  is  the  hypothenuse  of  a 
right  angled  triangle  whose  sides  are  P'M  and  P''M, 

Hence 

L  =  y/p'M  +  P^'  ;...(!) 
but  P'm'=  NC  =  NB  +  Bc'=  (OD  -  0A)2  +  (DC  -  AN)^  = 
(x"  -  x'Y  +  (y"  -  y')\  and  Pm'=  (P"C  -  P'N)^  = 
(."  -  zj. 


•.L=  V(a 


r  +  i:y"  -  y'f  +  {^"  -  ^7  •  •  •  (2) 


CO-ORDINATES.  221 

Cor.    If  x'  =  0,  1/  =  0,  z'  =  0,  then  the  point  P'  coincides 
with  the  origin  and 


.•.L  =  Vx''2  +  2/''^'  +  ,^''^  ...  (3) 
expresses  the  distance  of  a  point  from  the  origin. 

169.  Given  the  length  and  the  directional  angles  of  a  line 
joining  any  i^oint  tvith  the  origin  to  find  the  co-ordinates  of  the 
point. 

The  Directional  angles  of  a  line  are  the  angles  tvhich  the  line 
makes  with  the  co-ordinate  axes. 

Let  P  (x,  y,  z),  Pig.  75,  be  any  point,  then  OP  ^  L  will  be 
its  distance  from  the  origin.  Let  POX,  POY,  POZ  =  «,  ^,  y, 
respectively. 

Since  OM,  OlST,  OR  (=  x,  y,  z)  are  the  projections  of  OP 
on  X,  Y,  Z,  respectively,  we  have 

£c  =  L  cos  «    I 

y  =  L  cos  y8  I    ...  (1) 
s  =  L  cos  /    J 

for  the  required  co-ordinates. 

Cor.    Squaring  and  adding  equations  (1),  we  have 

^^  +  Z/^  +  «"  =  L^  (cos^  «  +  cos^  /3  +  cos^  y) ; 
but  ic^  +  2/2  _j_  ^2  _  L2  ^j.^_  jg3  (^3^) . 

hence  cos^  «  -\-  cos^  /3  +  cos^  y  =  1  .  .  .   (2) 

That  is,  the  sum  of  the  squares  of  the  directional  cosines  of  a 
space  line  is  equal  to  unity. 

ScHOL.  The  directional  angles  of  any  line,  as  PT",  Pig.  76, 
are  the  same  as  those  which  the  line  makes  with  three  lines 
drawn  through  P'  ||  to  X,  Y,  Z.  The  projections  of  PT''  on 
three  such  lines  are  x"  —  x',  if  —  ij ,  z"  —  z',  Art.  168 ;  hence 

x"  —  x'  =  L  cos  «  1 

/'  -  /  =  L  cos  ;8  j.    ...  (3) 

z"  —  z'  =  Jj  COS  y 


222  SOLID  ANALYTIC  GEOMETRY. 

EXAMPLES. 

Required  the  length  of  the  lines  joining  the  following 
points : 

1.  (1,  2,  3),  (- 2,  1,  1),     _    4.    (0,  0,  0),  (2,  0,  1). 

Ans.     V14.  Ans,     ^5. 

2.  (3,  -  2,  0),  (2,  3,  1).     _    5.    (0,  4,  1),  (-  2,  -  1,  -  2). 

Ans.     V27.  Ans.     V38. 

3.  (0,  3,  0),  (3,  -  1,  0).  6.    (1,  -  2,  3),  (3,  4,  6). 

Ans.    5.  Ans.    7. 

7.  Find  the  distance  of  the  point  (2,  4,  3)  from  the  origin  ; 
also  the  directional  cosines  of  the  line. 

8.  A  line  makes  equal  angles  with  the  co-ordinate  axes. 
What  are  its  directional  cosines  ? 

9.  Two  of  the  directional  cosines  of  a  line  are  Vf  and  ^ 
What  is  the  value  of  the  other  ? 

10.  If  (x'',  y',  z')  and  (x",  y",  z")  are  the  co-ordinates  of  the 
extremities  of  a  line  show  that 

(x'  +  x"     y'  +  y"     z'  +  ^-\ 
\      2       '        2       '        2      ) 
are  the  co-ordinates  of  its  middle  point. 

THE   POLAR   SYSTEM. 

170.  The  position  of  a  space  point  is  completely  determined 
when  we  know  its  distance  and  direction  from  some  fixed  point. 
For  a  complete  expression  of  the  direction  of  the  point  it  is 
necessary  that  two  angles  should  be  given.  The  angles 
usually  taken  are 

1st,  The  angle  which  the  line  joining  the  point  and  the 
fixed  point  makes  with  a  plane  passing  through  the  fixed 
point ;  and  2d,  The  angle  which  the  projection  of  the  line  join- 
ing the  points  on  that  plane  makes  with  a  fixed  line  in  the 
plane. 


CO-ORDINA  TES. 


223 


Fig.  77. 

Let  O  be  the  fixed  point  and  P  the  point  whose  position  we 
wish  to  determine.  Join  0  and  P,  and  let  XOY  be  any  plane 
passing  through  0.  Leb  OX  be  a  given  line  of  the  plane 
XOY.  Draw  PB  1  to  XOY  and  pass  the  plane  PBO  through 
PB  and  OP.  The  intersection  OB  of  this  plane  with  XOY 
will  be  the  projection  of  OP  on  XOY.  The  angles  POB  (6), 
BOX  (<p)  and  the  distance  OP  (r),  when  given  completely  de- 
termine the  position  of  P.  Por  the  angle  go  determines  the 
plane  POB,  the  angle  6  determines  the  line  OP  in  that  plane, 
and  the  distance  r  determines  the  point  P  on  that  line. 

This  method  of  locating  a  point  is  called  the  Polar  Sys- 
tem. The  angles  6  and  go  are  called  VECToraAL  Angles,  and 
the  distance  r  is  called  the  Radius  Vector  of  the  point. 
The  point  P,  when  written  (r,  6,  cp),  is  said  to  be  expressed  in 
terms  of  its  Polar  Co-ordinates. 

It  is  evident  by  giving  all  values  from  0  to  360°  to  0  and 
cp,  and  all  values  from  0  to  co  to  r  that  every  point  in  space 
may  be  located. 


171.    Given  the  polar  co-ordinates  of  a  point  to  find  its  rec- 
tangular co-ordinates. 

Draw  OY  1  to  OX  and  in  the  plane  BOX  ;  draw  OZ  1  to 


224  SOLID  ANALYTIC  GEOMETRY. 

OY  and  OX,  and  let  OX,  OY,  OZ  be  the  co-ordinate  axes. 
Draw  BA  ||  to  OY ;  then,  Fig.  77, 

(OA,  AB,  BP)  =  (x,  y,  z)  are  the  rectangular  co-ordinates  of  P. 
From  the  triangle  BOP,  we  have 

z  =  r  sin  B. 

From  the  triangle  ABO,  we  have 

X  =  OB  cos  (p. 

But  OB  =  r  cos  0  .-.  X  =  7-  cos  $  cos  (p. 
From  the  same  triangle  we  have 

y  =  OB  sin  cp, 
y  =  r  cos  6  sin  g). 
Henee  x  =  r  cos  0  cos  cp  1 

y  =  r  cos  0  sin  qo  V    ...  (1) 
z  =  r  sin  6 

express  the  required  relationship. 

CoR.  If  P  {x,  y,  z)  be  the  co-ordinates  of  any  point  on  a 
locus  whose  rectangular  equation  is  given  then  equations  (1) 
are  evidently  the  equations  of  transformation  from  a  rectangu- 
lar system  to  a  polar  system,  the  pole  being  coincident  with  the 
origin. 

Finding  the  values  of  r,  9  and  cp  from  (1)  in  terms  of  x  and 
y,  we  have 


r  =  Va;2  -\- f  -\-  ■«'  ' 
'  =  tan-^ 


Vx2  +  3/2    }    •  •  •  (2) 


tan-^  ^ 

X 


J 


for  the  equations  of  transformation  from  a  polar  system  to  a 
rectangular  system,  the  origin  and  pole  being  coincident. 


CO-ORDINA  TES.  225 

EXAMPLES. 

Find  the  polar  co-ordinates  of  the  following  points  : 

1.  (2,  1,  1).  .  3.    (10,  2,  8). 

2.  (V3,  1,  2  V3).  4.    (3,  -  1,  4). 
Find  the  rectansrular  co-ordinates  of  the  following : 


5. 


6. 


(5,  30°,  60°).  7.    (6,|,^V 


Find  the  polar  equations  of  the  following  surfaces,  the  pole 
and  origin  being  coincident: 

9.  x^  -\-  y-  -\-  z-  '=  a^.  Ans.    r  =  a. 

10.  z  -[-  sx  -\-  ty  —  c  =  0. 

Ans.     r  = . 

sm  0  -\-  s  cos  0  cos  (f  -\-t  cos  0  sin  qo 

Find  the  directional  cosines  of  the  lines  joining  the  follow- 
ing pairs  of  points : 

11.  (1,  2,  -  1),  (3,  2,  1).  13.    (2,  -  1,  -  5),  (4,  5,  6). 

12.  (4,  -  1,  2),  (-  1,  3,  2).         14.    (0,  2,  0),  (3,  0,  1). 

15.    If  (cc',  y',  z)  and  {x",  y",  z")  be  the  co-ordinates  of  two 
space  points,  show  that  the  point 

mx"  -\-  nx'      my"  -\-  ny'       mz"  +  nz' 
in  -\- 11  m  -{-  n  m  -\-  n 

divides  the  line  joining  them  into  two  parts  which  bear  to 
each  other  the  ratio  m  :  n. 


226. 


SOLID  ANALYTIC  GEOMETRY. 


CHAPTER    11. 


THE  PLANE. 


172.    To  deduce  the  equation  of  the  plane. 

Let  us  assume  as  the  basis  of  the  operation  the  following 
property  : 

If  on  a  perpendicular  to  a  p)lane  two  points  equidistant  from 
the  plane  he  taken,  then  every  point  in  the  plane  is  equidistant 
from  these  two  points,  and  any  point  not  in  the  plane  is  un- 
equally distant. 


Let  ABC  be  any  plane.  Draw  OR  -L  to  ABC,  and  meeting 
it  in  R.  Produce  OR  until  RR'  =  OR  =  p.  Every  point  in 
the  plane  is  equally  distant  from  0  and  R'.  Let  P  (cc,  y,^,) 
be  any  point  of  the  plane ;  let  OjST,  MN,  MR',  the  co-ordinates 


THE  PLANE.  227 

of  R'  =  d,  e,  f,  respectively.     Then  from  Art.  168,  (2),  we  have 
PR' =  (fZ  -  xy  +  (e  -  yf  +  {f-zy 
From  the  same  article,  equation  (3),  we  have 
OP-  =  a;2  +  2/'  +  «' ; 
hence,  by  the  assumed  property, 

(d  -  xy  +  (e  -  yy  +  (/  -  zy  =  x^J^y^^  z\ 
Simplifying  this  expression,  we  have 

dx  +  ey  +fz  =  d"+'"^+f"  ...   (1) 

for  the  required  equation. 

173.  To  find  the  equation  of  a  plane  in  terTns  of  the  per- 
pendicular^ to  it  from  the  origin  and  the  directional  cosines  of 
the  perpendicular. 

Let  «,  /3,  and  7  be  the  directional  angles  of  the  perpendicu- 
lar OP/ (=2^),  Fig.  78.  Since  ON,  MN,  MR'  (=  d,  e,  f)  = 
the  projections  of  OR'  on  the  co-ordinate  axes,  we  have  (Art. 
169,  (1)  ) 

d  =  2j9  cos  M  1 

e  =  2pcos  (3[    ...   (1) 

/=  2p  cos  y 


Substituting  these  values  in  (1),  Art.  172,  and  remembering 
that  cos^  w  -f-  cos^  (3  -\-  cos^  7  =  1,  we  have 

X  cos  a  -\-  y  cos  (3  -\-  z  cos  J  =  p  .  .  .  (2) 

for  the  required  equation.     Equation  (2)  is  called  the  Normal 
Equation  of  the  plane. 


cos  «  = 


Since     OR'  =  2p  =  ^d^  +  e^  +f%  equations  (1)  give 

d  e 

,  ,  cos  B  =  — ,  , 

f 

cos  y  ==     -  . 


228  SOLID  ANALYTIC  GEOMETRY. 


Substituting  these  values  in  (2),  we  have 
d  6 


■yjd'  ^e'  -^p  Vf/'  +  e^  +  / 

/ 

VfZ^  +  e^  +  /^  ^  =i'  •  •  •   (  ) 

for  the  equation  of  the  plane  expressed  in  terms  of  the  co-or- 
dinates of  a  point  on  the  perpendicular  to  it  from  the  origin 
and  the  perpendicular. 

Cor.  1.    If  ^:*  =  0  in  (2),  we  have 

X  cos  «  +  2/  cos  /?  +  s  cos  y  =  0  .  .  .  (4) 
for  the  equation  of  a  plane  through  the  origin. 
CoK.  2.    If  M  =  90°,  cos  «  =  0,  hence 

y  cos  ^  -\-  ^  cos  y  =  p  .  .  .  (5) 

is  the  equation  of  a  plane  -L  to  the  YZ-plane. 
If  ^  =  90°,  we  obtain  similarly 

X  cos  a  -\-  z  cos  '/  =  ^  .  .  .  (6) 

for  the  equation  of  a  plane  ±  to  the  XZ-plane. 

If  J.  =  90°,  then 

X  cos  «  +  2/  cos  p  ^p  .  .  .  (7) 
is  the  equation  of  a  plane  1  to  the  XY-plane. 

CoE.  3.    If  «  =  90°  and  (3  =  90°,  then 


P 


...  (8) 


cos  7 

is  the  equation  of  a  plane  _L  to  YZ  and  XZ,  and  hence  |(  to 
XY. 

Similarly,  we  find 

1/  =  -^—  •  •  •  (9) 
-^       cos  ^  ^  ^ 

X  =  -^^—  .  .  .  (10) 
cos  « 

for  the  equations  of  planes  ||  to  XZ  and  YZ  respectively. 


THE  PLANE. 


229 


Cor.  4.    If  p  =  0  in  (8),  (9),  and  (10),  then 
«  =  0 


2/  =  0 


(11) 


are  the  equations  of  XY,  XZ,  and  YZ,  respectively. 

174.  2h  find  the  equation  of  a  plane  in  terms  of  its  in- 
tercepts. 

Let,  Fig.  78,  OA  =a,  OB  =  5,  OC  =  c.  Since  OR  {=p)  is 
perpendicular  to  the  plane  ABC,  we  have  from  the  right  tri- 
angles ORA,  ORB,  and  ORC 


(a) 


Substituting   these    values   in   the    normal    equation    and 
reducino^,  we  have 


cos  «  = 

p    T 

a 

cosy3  = 

cos  y  = 

P 

c 

a        b        G 


(1) 


for  the  required  equation.     Equation  (1)  is  called  the  Sym- 
metrical Equation  of  the  plane. 

175.    Every  equation  of  tJie  first  degree  between  three  vari- 
ables represents  a  plane. 

The  most  general  equation  of    the   first  degree   between 
three  variables  is  of  the  form 

Ace  +  By  +  C«  =  D  .  .  .  (1) 


Dividing  both  members  of  this  equation  by  VA^  -\-W'  -{■  C^, 
we  have 

A  B  C  ^ 

VA2  +  B2  +  C^  ^  "^  VA^  _j_  B2  -j-  C^  ^  "^  VA2  4-  B2  +  C^ 
_  D 

VaJTb^Tc^  •  •  •  (2) 


230 


SOLID  ANALYTIC  GEOMETRY. 


Comparing  (2)  with  (3)  of  Art.  173,  we  see  that  the  co- 
efficients of  the  variables  are  the  directional  cosines  of  some 
line  expressed  in  terms  of  the  co-ordinates  of  one  of  its 
points,  and  that  the  second  member  measures  the  distance  of 
a  plane  from  the  origin ;  hence  (2)  and  therefore  (1)  is  the 
equation  of  a  plane. 

176.  To  find  the  equations  of  the  traces  and  the  values  of 
the  intercepts  of  a  plane  given  by  its  eq^uation. 


Fig.  79. 
Let  ABC  be  the  plane  and  let  its  equation  be 
Ace  +  By  -f  C^  =  D. 

1.    To  find  the  equations  of  the  traces  AB,  BC,  AC. 

The  traces  are  the  intersections  of  the  given  plane  with 
the  co-ordinate  planes  ;  hence,  combining  their  equations,  we 
have 

Ax  -f  By  +  C«  =  D  I  ._  ^^  _^  j5^  _  D.  Trace  on  XY  (AB)  ...  (1) 

Ao.  +  By  +  C^  =  D I  _._  ^^  ^  ^^^  _  j^    ^^^^g  ^^^  X2  (AC)  ...  (2) 
y  =  0  ) 

Ax  +  By  -f  C,^  =  D I  _.  g    ^  ^^  _  D.  Trace  on  YZ  (BC)  ...  (3) 

X  =  0  ) 


THE  PLANE.  231 

2.    To  find  the  intercepts  OA,  OB,  OC. 

The  points  A,  B,  C  are  the  intersections  of  the  given  plane 
with  the  co-ordinate  planes  taken  in  pairs  ;  hence,  combining 
their  equations,  we  have 

^  =  0  l.-.:.  =  ^=OA.  .  .  (4) 

2/  =  0  J 

Ax -{- By -{- Gz  =  B  ^ 

x  =  0  L-.y  =  -^=OB  ...  (5) 

I 

x  =  0  I.-.  ^=  —  =0C  .  .  .  (6) 

y  =  o  J  ^ 

Cor.  If  the  plane  is  perpendicular  to  XZ  its  Y-inter- 
cept  =  OB  :=  cc  ;  hence,  equation  (5),  B  =  0.  Making  B  ^  0 
in  the  general  equation,  we  have 

Aic  +  Cs  =  D  .  .  .  (7) 

But  (7)  and  (2)  are  the  same  equations ;  hence,  a  jJerpendic- 
ular  pjlane  and  its  trace  on  the  plane  to  tohich  it  is  2>&'>y^'n.dic- 
idar  have  the  same  equation. 

YJl.  If  X  cos  a  -\-  y  cos  jB  -\-  z  cos  y  ^  p  be  the  normal  equa- 
tion of  a  plane,  then  x  cos  u  -\-  y  cos  ft  -\-  z  cos  y  =  p  -^  d  is  the 
equation  of  a  parallel  plane  at  the  distance  d  from  it. 

For  the  directional  cosines  of  the  perpendiculars  are  the 
same ;  hence,  the  perpendiculars  are  coincident ;  hence,  the 
planes  are  parallel.  The  distance  of  the  planes  apart  is  equal 
to  the  difference  of  the  perpendiculars  drawn  to  them  from 
the  origin;  but  this  difference  is^  -|-  d—p\  i.e.,  -i-  d.  Hence, 
the  proposition. 

Cor.  If  (x' ,  y',  z')  be  a  point  in  the  plane  whose  distance 
from  the  origin  \q  p)  J^d;  then 

:^d  ^x'  cos  «  +  y'  cos  ^  +  z'  cos  y  —  p  ■  ■  ■  (1) 


232  SOLID  ANALYTIC  GEOMETRY. 

is  its  distance  from  the  parallel  plane  whose  distance  from 
the  origin  is  p.     From  equations  {ci),  Art.  174,  we  have 

cos  «  =  i-  ^  cos  13  =  ^ ,  cos  5'  =  =L  ; 
a  b  c 

2  2  2 

hence        cos^ «  +  cos'-^  ft  -\~  cos^  y  =  -2L  -)-  E — [_  P—  —  i. 

a^        IP-         c^ 

These  values  in  (1)  give 

for  the  expression  of  the  distance  of  a  point  from  a  plane 
which  is  given  in  its  symmetrical  form. 

Let  the  student  show  that  the  expression  for  d  becomes 
^^^AaM;E£^C£^-D      _ 

VA--^  +  B^  +  C^ 

when  the  equation  of  the  plane  is  given  in  its  general  form. 

What  is  the  significance  of  the  double  sign  in  (1),  (2),  and 
(3)? 

178.  To  find  the  equation  of  a  i^lane  which  jpasses  through 
three  given  lyoints. 

Let  {x',  tj',  z'),  {x",  y",  z"),  {x"' ,  y"\  z"')  be  the  given  points. 
Since  the  equation  we  seek  is  that  of  a  plane,  it  must  be 

Aaj  +  B^/  +  C,^  =  D  .  .  .  (1) 
in  which  A,  B,  C,  D  are  to  be  determined  by  the  conditions 
imposed. 

Since  the  plane  is  to  contain  the  three  given  points,  the  co- 
ordinates of  each  of  these  must  satisfy  its  equation ;  hence, 
the  following  equations  of  condition  : 

Kx'  +  Bt/'  +  Cs'  =  D 
Kx"  +  By''  +  C;s''  =  D 
Aa;'"  +  B?/'"  +  C.~"'  =  D. 

These  three  equations  contain  the  four  unknown  quantities 
A,  B,  C,  D.     If  we  find  from  the  equations  the  values  of  A, 


THE  PLANE.  233 

B,  C  in  terms  of  D  and  the  known  quantities,  and  substitute 
these  vahies  in  (1),  each  term  of  the  resulting  equation  will 
contain  D  as  a  factor.     Let 

A  =  A'D,  B  =  B'D,  C  =  CD  be  the  values  found. 
Substituting  in  (1),  we  have 

MT>x  +  B'Dy  +  C'D«  =  D. 
...  A'cc+BV  +  C'a  =  l  •  •  •  (2) 
is  the  required  equation. 

179.  The  preceding  discussion  has  elicited  the  fact  that 
every  equation  of  the  first  degree  between  three  variables 
represents  a  plane  surface.  It  remains  to  be  shown  that  every 
equation  between  three  variables  represents  a  surface  of  some 
kind. 

Let  ^=f{x,  y)  ■  ■  '  (1) 

be  any  equation  between  the  three  variables  (x,  y,  z).  Since 
X  and  y  are  independent,  we  may  give  them  an  infinite  number 
of  values.  For  every  pair  of  values  thus  assumed  there  is  a 
point  on  the  XY  plane.  These  values  in  (1)  give  the  corre- 
sponding value  or  values  of  z,  which,  laid  off  on  the  perpen- 
dicular erected  at  the  point  in  the  XY  plane,  will  locate  one 
or  more  points  on  the  locus  of  the  equation.  But  the  number 
of  values  of  z  for  any  assumed  pair  of  values  of  x  and  y  are 
necessarily  finite,  while  the  number  of  pairs  of  values  which 
may  be  given  x  and  y  are  infinite  ;  hence  (1)  must  represent 
a  surface  of  some  kind. 
If 

^=f{^, y)\  ,  ,  ,  (2) 

z  =  (p{x,y)\ 

be  the  equations  of  two  surfaces,  then  they  will  represent  their 
line  of  intersection  if  taken  siviultaneously.  For  these  equa- 
tions can  only  be  satisfied  at  the  same  time  by  the  co-ordinates 
of  points  common  to  both.  Hence,  in  general,  two  equations 
between  three  variables  determine  the  position  of  a  line  in  sp>ac6. 


234  SOLID  ANALYTIC  GEOMETRY. 

If  z  ^f{x,  y)    1 

z  =  ^{x,ij)\    .  .  .   (3) 

z-=xp{x,  y)  j 
be  the  equations  of  three  surfaces,  then  they  will  represent 
their  point  or  points  of  intersection  when  considered  as  simul- 
taneous.    Hence,    in    general,    three    equations   between   three 
variables  determine  the,  positions  of  space  points. 

EXAMPLES. 

Find  the  traces  and  intercepts  of  the  following  planes: 

1.  X  -2y  +  z  =Q.  6.    -  -  ^  +  -  =  1. 

■^  2        3       4 

2.  |.-3/+|=l.  7.   |_|-|  =  1. 

3.  .-,  +  4.,=  l.  8.   2_^-^f  +  ^  =  l. 

4.  2cc  +  32/-4,^=0.  9.   ^  +  |_^  =  1. 

5    ^  —  ^  JL.  y  —  ^  =  2  10?^  —  ^  =  ^ 

2^3  ■  -     y       ^       4:' 

11.  The  directional  cosines  of  a  perpendicular  let  fall  from 

2  12 
the  origin  on  a  plane  are  - ,  - ,  -;  required  the  equation  of  the 

o  o  o 
plane,  the  length  of  the  perpendicular  =  4. 

Ans.     ^  +  i^  +  A  =  1. 
6       12        6 

Required  the  equations  of  the  plane  whose  intercepts  are 

as  follows  : 

12.  1,  2,  3.  14.    ^,  |,  -2. 

13.  2,   -  1,  3.  15.    _  1,  _  I ,   _  4. 

o 

16.    What  is  the  equation  of  the  plane,  the  equations  of 
whose  traces  are  x  —  3  ^z  =  4  and  cc  -|"  ^  =  4  ? 

Ans.   x  —  3?/4-«  =  4. 


THE  PLANE.  235 

17.  The  co-ordinates  of  the  projection  of  a  point  in  the 
plane  cc  —  3?/  +  2s;  =  2on  the  XY  plane  are  (2,  1) ;  required 
the  distance  of  the  point  from  the  XY  plane. 

Ans.    §. 

Write  tlie  equations  of  the  planes  which  contain  the  follow- 
ing points : 

18.  (1,  2,  3),  (-  1,  2,  -  1),  (3,  2,  0). 

19.  (4,  1,  0),  (2,  0,  0),  (0,  1,  2). 

20.  (0,  2,  0),  (3,  2,  1),  (-  1,  0,  2). 

21.  (2,  2,  2),  (3,  3,  3),  (-  1,  -  1,  -  1). 

Find  the  point  of  intersection  of  the  planes 

22.  2x-\-y  —  z  =  ^.  23.   2x  —ij  +  z  =  10. 
2x  —  3z  -\-ij  =  10.  x-{-y-2z  =  3. 

X  -\-  y  —  «=2.  2  X  —  4?/-l-5s  =  6. 

24.   2x  —  y-z  =  2. 
2x  —  3y-{-z  =  10. 
2x—y-{-2z  =  8. 

Find  the  distance  of  the  point  (2,  1,  3),  from  each  of  the 
planes 

25.  x  cos  60°  +  y  cos  60°  +  z  cos  45°  =  9. 

26.  x-\-3y-z  =  8. 

27.   x+^  -\-3z  =  4..  28.  -  -  ^  +  -  =  1. 

2  3   2   5 

29.  Find  the  equation  of  the  plane  which  contains  the 
point  (3,  2,  2)  and  is  parallel  to  the  plane  x  —  2  y  -\-  z  =  6. 

Reduce  the  following  equations  to  their  normal  and  sym- 
metrical forms  : 

30.  2x-3y  -\-z  =  4.  31.   4.x  -^  2y  —  z  =  ~  . 

32.    ~x-\-y  —  ~z  =  ^. 
3  "^       4 

33.  If  s,  /,  s"  represent  the  sides  of  the  triangle  formed  by 
the  traces  of  a  plane,  and  a,  h,  c  represent  the  intercepts, 
show  that  s2  4.  s'^  4-  s'""  ^2(a^  -^b^  +  c"). 


236 


SOLID  ANALYTIC  GEOMETRY. 


CHAPTER    III. 


THE   STRAIGHT   LINE. 


180.    To  deduce  the  equations  of  the  straight  line. 

The  straight  line  in  space  is  determined  when  two  planes 
which  intersect  in  that  line  are  given.  (See  Art.  179.)  The 
equations  of  any  two  planes,  therefore,  may  be  considered  as 
representing  a  space  line  when  taken  simultaneously.  Of  the 
infinite  number  of  pairs  of  planes  which  intersect  in  and  de- 
termine a  space  line,  two  of  its  projecting  planes  —  that  is, 
two  planes  which  pass  through  the  line  and  are  perpendicular 
to  two  of  the  co-ordinate  planes  — give  the  simplest  equations. 
For  this  reason  two  of  these  planes  are  usually  selected. 


Fig.  so. 


THE  STRAIGHT  LINE.  237 

Let  PBM  be  the  plane  which  projects  a  space  line  on  XZ, 
then  its  equation  will  be  of  the  form 

X  ^=  sz  -{-  a  (see  Art.  176,  Cor.) 
in  which  s  =  tan  ZBP  and  a  =  OA. 

Let  P'B'M'  be  the  plane  which  projects  the  line  on  YZ, 
then  its  equation  will  be 

y=tz+b, 
in  which  t  =  tan   ZBT'  and  b  =  OA'. 

But  the  tAvo  planes  determine  the  line ;  hence 
X  =  sz  -\-  a  )^  ,-.s 

y  =  tz  +  b\     '    '    '    ^^ 

are  the  required  equations. 

Cob.  1.    If  a  =  0  and  5  =  0,  then 

X  =  sz^    _  _  _   ^2') 
y  =  tz) 
are  the  equations  of  a  line  which  pass  through  the  origin. 
Cob.  2.    If  s  =  0  and  ^^  =  0,  we  have 

^  =  ''1    ...   (3) 
y  =  bS 

for  the  equation  of  a  line  ||  to  the  Z-axis. 

CoR.  3.  Since  equations  (1)  express  the  relation  existing 
between  the  co-ordinates  of  every  point  on  the  space  line,  if 
we  eliminate  z  from  these  equations  we  obtain  the  immediate 
relation  existing  between  x  and  y  for  points  of  the  line.  But 
this  relation  is  evidently  the  same  for  all  points  in  the  pro- 
jecting plane  of  the  line  which  is  L  to  XY  and  therefore  for 
its  trace  on  XY.  But  the  trace  is  the  projection  of  the  line 
on  XY ;  hence,  eliminating,  we  have 

sy  —  tx  =  bs  —  at  .  .  .   (4) 
for  the  equation  of  the  projection  of  the  line  on  XY. 

181.  We  have  found,  Art.  169,  SchoL,  for  the  length  of  a 
line  joining  two  points  the  expression 

L  ^  x"  -x'  ^   if  -  if    ^  z"  -  z' 
cos «  cos  ^  cos  y 


238  SOLID  ANALYTIC  GEOMETRY. 

Eliminating  L  and  letting  x",  y",  z"  {=  x,  y,  z)  be  the  co- 
ordinates of  any  point  on  the  line,  we  have 


X  —  X   y  —  y   z  —  z 


...  (1) 


cos  «  COS  ^  COS  7 

for  the  Symmetrical  Equation  of  a  straight  line. 

182.    To  find  ivhere  a   line  given  by  the  equations  of  its 
projections  pierces  the  co-ordinate  planes. 

CT    —    ^^      I      ft    f 

Let  ^"^  T  7    (he  the  equations  of  the  line. 

y  =  tz  -\-b  ^  -L 

1.  To  find  where  the  li7ie  jnerces  the  XY-plane. 
The  equation  of  the  XY-plane  is 

z  =  Q. 
Since  the  jDoint  of  intersection  is  common  to  both  the  line 
and  the  plane,  its  co-ordinates  must  satisfy  their  equations. 
Hence 

X  ^=  sz  -\-  a 

y  =  tz  -\-b 

z  =  0 
are  simultaneous  equations.     So  treating  them  we  find 

{a,  b,  0) 
to  be  the  required  point. 

2.  To  find  tvhere  the  line  p)ieTces  the  XZ-plane. 
The  equation  of  the  XZ-plane  is 

Combining  this  with  the  equations  of  the  line,  we  have 
at  —  ■^^    A         ^ 


t  t 

for  the  required  point. 

3.    To  find  xohere  the  line  pierces  the  YZ-plane. 

X  ^  sz  -\-  a') 

y  =  tz  -\-  b  >  are  simultaneous ; 

x  =  0  ) 

hence  [0,  ^^  ~  ^^         ^ 


s  s 

is  the  required  point. 


THE  STRAIGHT  LINE.  239 

183.  To  find  the  equations  of  a  line  passing  through  a  given 
point. 

Let  (x\  y',  z')  be  the  given  point. 

Since  the  line  is  straight  its  equations  are 

X  =  sz  +  a  I  .-|^^ 

y  =  tz  +  b^ 

in  which  the  constants  are  unknown. 

Since  it  is  to  pass  tlirougii  the  point  (x',  y',z')  its  equations 
must  be  satisfied  for  the  co-ordinates  of  this  point ;  hence 
the  equations  of  condition : 

X  =  sz  -\-  a  I  /9^ 

y'    =    tz'    +h     \         '      '      '      ^'^^ 

As  the  three  conditions  imposed  by  these  four  equations 
cannot,  in  general,  be  fulfilled  by  a  straight  line,  we  must 
eliminate  one  of  them.  Subtracting  the  first  equation  in 
group  (2)  from  the  first  in  group  (1)  and  the  second  in  group 
(2)  from  the  second  in  group  (1),  we  have 

iC  CC    =  S  (Z  ^  /  )  f'X\ 

y-y'  =  t{z-z')S     '  -  '  ^^ 

for  the  general  equations  of  a  straight  line  p)Cissing  through  a 
point. 

184.  To  find  the  equations  of  a  line  passing  through  two 
given  p)oi7its. 

Let  (x',  y',  z'),  {x'\  y",  z")  be  the  given  points. 
As  the  line  is  straight  its  equations  are 

y  =  tz-{-b\'     '    '    '    ^   ^ 

in  which  the  constants  are  to  be  determined. 

As  it  is  to  pass  through  (x',  y' ,  z'),  we  must  have 

x'  =  sz'  -\-  a 


(2) 
y'  =  tz'  -^b  '  ^  ^ 


240 


SOLID  ANALYTIC   GEOMETRY. 


As  it  is  to  pass  through  (x",  y",  z"),  we  must  have  also 
x"  =  sz"  +  a  ~>  /ON 

y'^  =  tz"  -\-l\ 

As  these  six  equations  impose  four  conditions  on  the  line, 
we  must  eliminate  two  of  them.  The  conditions  of  the 
proposition,  however,  require  the  line  to  pass  through  the 
two  points ;  hence  we  must  eliminate  the  other  two. 

Elimiting  a  and  h  from  groups  (1)  and  (2),  by  subtraction, 
we  have 

X  —  x'  =  s  {z  —  z'")^ 
y-y'^t{z-z')\    ■  •  •   ^^) 

Now,  eliminating  a  and  b  from  (2)  and  (3),  we  have 

x'  —  x"  =  S  {z'  —  «")  )  /f-N. 


^J  -y"  =  t  {z'  -  z")  i 
Eliminating  s  and  t  between  (4)  and  (5),  we  have 


x  =  (z  —  z) 


y-y'-^-f^,i?-^') 


for  the  required  equations. 


(6) 


EXAMPLES. 

1.  Given  the  line  ^,  ^^  4  ^  _  3  [•  required  the  equation  of 

the  projection  on  XY. 

Ans.     2  X  —  y  =  5. 

2.  How  are  the  following  lines  situated  with  reference  to 
the  axes  ? 

x=2}      y=Ol     2/  =  01 


=S} 


Find  the  co-ordinates  of  the  points  in  which  the  following 
lines  pierce  the  co-ordinate  planes  : 


X  =  3z  — 
y  =  2z  + 


1} 


A       X  =    —  Z  —1\ 

*•    2/  =  2^  +  3  ) 


THE  STRAIGHT  LINE.  241 

6.    Given  (2,  1,  -  2),  (3,  0,  2)  ;  required 

(a)    The  length  of  the  line  joining  the  points. 

(h)    The  equation  of  the  line. 

(c)  The  points  in  which  the  line  pierces  the  co-ordinate 
planes. 

Find  the  equations  of  the  lines  which  pass  through  the 
points : 

7.  (2,  1,  3),  (3,  -  1,  - 1).         9.  (2,  -  1,  0),  (3,  0,  0). 

8.  (  -  1,  2, 3),  (  -  1,  0, 2).   10.  (1,  -  1,  -  2),  (-  1,  -2,  -  3). 

11.  The  projections  of  a  line  on  XZ  and  YZ  make  angles 
of  45°  and  30°  respectively  with  the  Z-axis,  and  the  line  in 
space  contains  the  point  (1,  2,  3) ;  required  the  equations  of 
the  line. 

X  =  z  —  2. 

12.  The  vertices  of  a  triangle  are  (2,  1,  3),  (3,  0,  —  1), 
(—2,  4,  3)  ;  required  the  equations  of  its  sides. 

13.  Is  the  point  (2,  —  1,  3)  on  the  line  which  passes  through 
(-  1,  3,  2),  (3,  2,  -  2)  ? 

14.  Write  the  equations  of  a  line  which  lies  in  the  plane 
x-2y  +  3z  =  l. 

Note.  —  Assume  two  points  in  the  plane;  the  line  joining 
them  will  be  a  line  of  the  plane. 

15.  Find  the  equation  of  a  line  through  (1,  —  2,  2)  which 
is  parallel  to  the  plane  x  —  y  -\-  z  =  4:. 

T  J-  2  s;  =  3        ^ 

16.  Find   the   point   in    which   the   line       _  ^    t   o  _  a  r 

pierces  the  plane  Sx-{-2y  —  z  =  4:. 

17.  Required  the  equation  of  the  plane  which  contains  the 
^  T  x-2z-l=0}  .  x-z-5  =  0  I 
t^^°l^^^^  2/-2^-2  =  0|  ^^'^  2/-4^  +  6  =  0      j 


242  SOLID  ANALYTIC  GEOMETRY. 

18.  Find  the  point  of  intersection  of  the  planes 

19.  Find  the  equations  of  the  projecting  planes  of  the  line 

2x  ^3y  —  z=  Q\- 

20.  Which  angles  do  the  following  planes  cross  ? 
X—  y-\-z  =  4:,  2x-\-y  —  3z  =  2,  X— 2y  —  z  =  l. 

185.    To  find  the  intersection  of  two   lines  given  hy  their 

ec[uations. 

_    .  X  ^  sz  -\-  a^        :,  X  =  s'z  -\-  a' 

Let  ^     ;   7  ^  and  .,        ,, 

y  =^tz  -\-b   '^  y  =  tz  -{-  0 

be  the  given  equations.  Since  the  point  of  intersection  is  com- 
mon to  both  lines,  its  co-ordinates  must  satisfy  their  equations. 
Hence  these  equations  are  simultaneous.  But  we  observe  that 
there  are  four  equations  and  only  three  unknown  quantities ; 
hence,  in  order  that  these  equations  may  consist  (and  the  lines 
intersect),  a  certain  relationship  must  exist  between  the  con- 
stants which  enter  into  them.  To  find  this  relationship,  we 
eliminate  x  between  the  fii^st  and  third,  y  between  the  second 
and  fourth,  and  z  between  the  two  equations  which  result. 
We  thus  obtain 

(s  -  s')  (b  -  b')  -  (t-  t')  (a  —  a')  =  0 

for  the  required  equation  of  condition  that  the  two  lines  shall 
intersect.  If  this  condition  is  satisfied  for  any  pair  of  as- 
sumed lines  the  lines  will  intersect,  and  we  obtain  the 
co-ordinates  of  this  point  by  treating  any  three  of  the  four 
equations  which  represent  them  as  simultaneous.  So  treating 
the  first,  second,  and  third  we  obtain 

s«'  —  s'a     .a'  —  a    ,    -,    a'  —  a 

— — :^ '  ^  z — J  +  ^'  - — V 


for  the  co-ordinates  of  the  required  point. 

ISToTE.  —  We  were  prepared  to  expect  that  our  analysis 
would  lead  to  some  conditional  equation,  for  in  assuming  the 
equations  of  two  space  lines  it  would  be  an  exceptional  case 


THE  STRAIGHT  LINE. 


243 


if  we  so  assumed  them  that  the  lines  which  they  represent 
intersected.  Lines  may  cross  each  other  under  any  angle  in 
space  without  intersecting.  In  a  plane,  however,  all  lines 
except  parallel  lines  must  intersect.  Hence,  no  conditional 
equation  arose  in  their  discussion, 

186.  To  find  the  angle  between  two  lines,  given  by  their 
equations,  in  terms  of  functions  of  the  angles  which  the  lines 
make  with  the  axes. 


Let 


sz  -\-  a 
tz^b 


,  a;  =  s'z  4-  »' 

and  ,,         ,, 

y  =  tz^b 


''} 


be  the  equations  of  the  two  lines.  The  angle  under  which 
two  space  lines  cross  each  other  is  measured  by  the  angle 
formed  by  two  lines  drawn  through  some  point  parallel  to 
their  directions. 


Let  OB  and  OC  be  two  lines  drawn  through   the  origin 
parallel  to  the  given  lines.     Then 

^  =  f|  and^=^. 
y  =  tz\  y  =  tz 

will  be  their  equations.     The  angle  between  these  lines  is  the 
angle  sought.     Let  go  (=  BOC)  be  this  angle. 


244  SOLID  ANALYTIC   GEOMETRY. 

Let  a',  p',  y'  represent  the  angles  which  the  line  BO  makes 
with  X,  Y,  Z,  respectively;  and  u",  (3'',  y"  the  angle  which 
CO  makes  with  the  same  axes.  Take  any  point  P'  (x',  y' ,  z') 
on  OB  and  any  point  V  (x",  y",  z")  on  CO  and  join  them  by 
a  right  line  forming  the  triangle  P'OP". 

Let  OP'  =  V,  OV"^  V,  and  PT"  =  L. 

Prom  the  triangle  P'OP'',  we  have 

L'2  +  L''2_L2  ,,. 

cos  cp  =  y^TjTT •    ■    •    (1) 

But  Art.  168,  equation  (3)  and  (2) 

L'2  =  x''-  +  y'-  +  z'% 
L"2  =  x"^^  y"'  +  z"% 
U  =   (x"  -  x'f  +   {y"  -  yj  +  iz!'  -  z'f 
=  x"^  +  2/"2  ^  ^//2  ^  ^'2  ^  ^/2  +  ~'2  _  2  (a;V  +  yY  +  «'0- 

Substituting  these  values  in  (1),  we  have 

x'x"  +  yY  +  «'^''  /o\ 

cos  q>  =  ^£,£,,^ •  •  •  (2) 

But  Art.  169,  (1) 

x'  =  L'  cos  a',  ?/'  =  L'  cos  f3',  z'  =  L'  cos  y' 
x"  =  V  cos  a",  y"  =  L"  cos  (3'',  z"  =  V  cos  y" 

Substituting  in  (2),  we  have 

cos  qo  =  cos  a'  cos  a"  +  COS  /3'  COS  yS"  +  cos  "/  cos  /"  .  .  .  (3) 
for  the  required  relation. 
Cor.    If  (p  =  90° 

cos  a'  COS  a"  -\-  COS  /5'  COS  ^"  +  COS  y'  COS  y"  =  0  .  .  .  (4) 

187.  To  find  the  angle  which  two  space  lines  Tnake  with  each 
other  in  terms  of  functions  of  the  angles  which  the  projections 
of  the  lines  'make  with  the  co-ordinate  axes. 

T  ^i_  X  ^^  SZ  f  -I   X  —  s  z 

Let  ,    y  and  ., 

y  =  tz\,  y  =  tz 

be,  as  in  the  preceding  article,  the  equations  of  the  lines 


THE  STRAIGHT  LINE.  245 

drawn  through  the  origin  parallel  to  the  given  lines.     Since 
P'  (x',  y',  z'),  Fig.  81,  is  a  point  on  the  first  line,  we  have 

x'  =  sz' 


'2  _l_  V2 


and,  Art.  168,     L"-^  =  x"^  +  tj'^  +  s' 

Eliminating,  we  find 

sh'  ,  tV  ,  \! 

y 


•^  -  Vi  +  s--^  +  ^^  VI +5'  +  ^-  Vi  +  s^  +  ^^' 

and  since  P"  {%" ,  y",  z")  is  a  point  on  the  second  line,  we 
have 

x"  =  s'z" 

y"  =  t'z", 

and,  Art.  168,     L"^  =  x"'-  +  y"^  +  s"^. 
Hence, 


y 


Vl  +  s'2  +  ^'2  V 1  +  s'-'  +  t"'  Vl  +  s'"  +  t'^ 

But,  Art.  169, 

f          0(1                            S                                 f/           00  s 

cos  «    =  =:  ^^=^z=^  ,  cos  (X      =  — -  =  z=z3^^=- 

L'       Vl+6'-'  +  ^'  ^         Vl  +  s'2  +  ^'' 

o,  y      t or,<i  R."  y       ^ 

,         «'  1  ,,        s"  1 

cos  7   =  — 7  =  —  —  ,  cos  7 


Substituting   these  values    in    equation  (3),  Art.  186,  and 
reducing,  we  have 

cos  go  =  -j-  —  ^— '  ...  (1) 

Vl  +  s^  +  ^2  Vl  +  s'2  _^  2^/2 

for  the  required  expression. 


246  SOLID  ANALYTIC  GEOMETRY. 

Cob.  1.    li  cp  =  0,  the  lines   are  parallel  and  equation    (1) 

becomes 

^  _ 1  +  ss'  4-  tf 

~       Vl  +  s^  +  f"    Vl  +  s'^  +  t'^  ' 

Clearing  of  fractions  and  squaring,  we  have 
(1  +  ^2  _^  ^2)  (1  +  ,^2  _^  ^^2)  _  (1  _^  ^,/  _j_  tty^ 

Performing  the  operations  indicated,  transposing  and  col- 
lecting, we  have 

(5'  _  s)2  _|_  (t'  _  ^)2  _^  (^5^'  _  s't^2  _  0. 

But  the  sum  of  the  squares  of  these  quantities  cannot  be 
equal  to  zero  unless  each  separately  is  equal  to  zero ;  hence 

s'  =  s,  f  =  t,  St'  =  s't  .  .  .  (2) 
are  the  conditions  for  parallelism  of  space  lines.  The  first 
two  of  these  conditions  show  that  if  two  lines  in  space  are 
parallel,  then  their  projections  on  the  co-ordinate  planes  are 
parallel  also.  The  third  condition  (st'  =  s't)  is  a  mere  conse- 
quence of  the  other  two,  and  may  be  omitted  in  stating  the 
conditions  for  parallelism. 

CoR.  2.  If  go  =  90°,  the  lines  are  perpendicular  to  each 
other,  and  equation  (1)  becomes 

^  ^  1  +  ss'  +  tif . 

Vl  -f  s^  +  f  Vl  +s'^  +  t'^  ' 
hence  1  +  ss'  +  tt'  =  0  .  .  .  (3) 

is  the  condition   for  perpendicularity  in  space. 

188.  Since  the  angle  which  a  line  makes  with  any  one  of 
the  co-ordinate  axes  is  the  complement  of  the  angle  which  the 
line  makes  with  the  co-ordinate  plane  to  which  that  axis  is 
perpendicular  if  Ave  let  «,  ^,  7  be  the-  complements  of  «',  /?',  y', 
respectively,  we  have 

S  .       o  t 

sm  a  = —  ,  sm  /3  = 


Vl  -}-  s^  ^e  Vl  +  s2  +  e 

sin  7  =  ——         — =.  ...   (1) 

for  the  sines  of  the  angles  which  a  space  line  makes  with  the 
co-ordinate  planes. 


THE  STRAIGHT  LINE. 


247 


TRANSFORMATION   OP   CO-ORDINATES. 

189.  To  find  the  equations  of  transformation  from  one 
system  of  co-ordinates  to  a  parallel  system,  the  origin  being 
changed. 


,/ 


~7^ 


Yr 


Fig.  82. 


Let  X,  Y,  Z  be  the  old  axes  and  X',  Y',  Z'  the  new. 

Let  P  be  any  point  on  the  locus  CM.  Draw  PB,  A'E, 
O'L  II  to  OZ  and  meeting  XOY  in  B,  E,  and  L.  Draw  BR 
and  produce  it  to  A ;  BE  will  be  ||  to  OY ;  draw  LN  ||  to  BE 
and  LE  ||  to  OX.  Then  (OA,  AB,  BP)  =  {x,  tj,  z)  are  the 
old  co-ordinates  of  the  point  P. 

(O'A',  A'B;  B'P)  =  {x',  y',  z')  are  the  new  co-ordinates  of 
the  point  P. 

(OK,  iSTL,  LO')  =  (a,  h,  c)  are  the  old  co-ordinates  of  the 
new  origin  0'. 

Prom  the  fisjure 


248 


SOLID  ANALYTIC  GEOMETRY. 


OA  =  ON  +  O'A',  AB  =  XL  +  A'B;  BP  =  LO'  +  B'P ; 
hence  x  =^  a  -{-  x',y  ^  b  -{-  y',  z  =  c  -\-  z' 

are  the  required  equations. 

190.  To  find  the  eqiiations  of  transformation  from  a  rec- 
tangular system  in  sjjace  to  an  oblique  system,  the  origin  being 
the  same. 


Fig.  83. 

Let  OX,  OY,  OZ  be  the  old  axes,  and  OX',  OY',  OZ'  the 
new. 

Let  «',  fi',  y'  be  the  angles  which  OX'  makes  with  OX,  OY, 
OZ  respectively. 

Let  a",  13",  f  be  the  angles  which  OY'  makes  with  OX,  OY, 
OZ  respectively. 

Let  (/",  13'",  Y"  be  the  angles  which  OZ'  makes  with  OX,  OY, 
OZ  respectively. 

Let  P  be  any  point  on  the  locus  CM.     Draw  PB  and  PB' 


THE  STRAIGHT  LINE.  249 

II  to  OZ  and  07/,  respectively,  and  let  B  and  B'  be  the  points 
in  which  these  lines  pierce  the  planes  XOY  and  X'OY'. 
Draw  B'A'  ||  to  OY' ;  then 

(OA,  AB,  BP)  =  (x,  y,  z)  are  the  old  co-ordinates  of  the 
point  P. 

(OA',  A'B',  B'P)  =  (x',  y',  z')  are  the  new  co-ordinates  of  the 
point  P. 

From  P,  B'  and  A'  let  fall  the  perpendiculars  PA,  B'D,  A'L 
on  the  X-axis ;  then  from  the  figure,  we  have 
OA  =  OL  +  LD  +  DA 

But  OL,  LD  and  DA  are  the  projections  of  OA',  A'B',  and 
PB',  respectively  on  the  X-axis,  and  each,  therefore,  is  equal 
to  the  line  whose  projection  it  is  into  the  cosine  of  the  angle 
'  which  that  line  makes  with  the  X-axis.     (See  Art.  169  (1)  .) 

.-.  OL  =  OA'  cos  «',  LD  =  A'B'  cos  «",  DA  =  B'P  cos  «'" 

i.e.,  OL  =  x'  cos  a',  LD  =  y'  cos  «",  DA  =  z'  cos  «'"; 
hence,  substituting,  we  have 

X  =  x'  cos  a'  +  y'  COS  «"  -\-  z'  COS  «'"      | 

Similarly  y  =  x'  cos  /3'  +  /  cos  y8"  +  5;' cosj8"'    I    ...  (1) 

Z  ^=  x'  cos  7'  -j-  if  cos  )"  -\-  z'  COS  7'" 

Of  the  nine  angles  involved  in  these  equations,  six  only  are 
independent,  for  since  the  old  axes  are  rectangular,  we  must 
have  (See  Art.  169,  equation  (2)  ). 

COS^  a'  -|-  COS^  ^'  4-  COS^  '/  =  1 

cos^  «"  +  cos2  ^"  +  cos^  f  =1      I    •  •  •  (2) 

C0S2  a'"  +  COS^  ^"'  +  C0S2  7'"  =  1  J 

Cor.  1.  If  we  suppose  the  new  axes  to  be  rectangular  also 
we  must  have  in  addition  to  equation  (2)  the  following  condi- 
tional equations  :  See  Art.  186,  Cor. 

cos  «'  cos  a"  +  cos  /3'  COS  /3"  -\-  cos  7'  cos  7"  =  0 

cos  «'  cos  «"'  -f  cos  /3'  cos  /3"'  +  cos  7'  cos  7'"  =  0      I    •  •  •  (^) 

cos  «"  cos  «'"  +  cos  y8"  COS  y8"'  +  cos  7"  cos  7'"  =  0  J 

Hence,  in  this  case,  only  three  of  the  nine  angles  involved 
in  equation  (1)  are  independent. 


250 


SOLID  ANALYTIC  GEOMETRY. 


THE    CONIC    SECTIONS. 

191.  The  Conic  Sections,  or,  more  simply,  The  Conics, 
are  the  curves  cut  from  the  surface  of  a  right  circular  cone  by 
a  plane. 

We  wish  to  show  that  every  such  section  is  an  ellipse,  a 
parabola,  an  hyperbola,  or  one  of  their  limiting  cases.  Art. 
146. 

192.  To  deduce  the  equation  of  the  conic  surface. 


Fig.  84. 

Let  CAEA'C  be  the  conic  surface,  generated  by  revolving 
the  element  CA  about  OZ  as  an  axis.  Let  P  be  any  point  on 
any  element  as  CE ;  let  OC  =  c  and  OEC  =  6. 

Draw  DP  ||  to  XY-plane  and  intersecting  OZ  in  D ;  draw 
PK  II  to  OZ,  KB  II  to  OY,  and  join  0  and  K  producing  it  to 
meet  the  base  circle  in  E. 

Then  (OB,  BK,  KP)  =  (x,  y,  z)  are  the  co-ordinates  of  P. 

From  the  similar  triangles  COE,  CDP,  we  have 

DC         OC        ,        .  ,^. 

=  tan  ^  .  .  .  (1) 


DP 


OE 


THE  STRAIGHT  LINE. 


251 


But  DC  =  OC  -  PK  =  c  -  z,  and  DP  =  OK  =  -s/x"  +  3/^; 
hence, 

c  —  z 


tan  6; 


i.e.,  (c  -  ,-5)-  =  (a;'  +  y^)  tan ' 

is  the  required  equation. 


(2) 


193.    To  find  the  equation  of  the  intersection  of  a  right  cir- 
cular cone  and  a  ])lane. 


Fig.  85. 


Let  GALA'  be  the  cone  and  X'OY  the  cutting  plane.  Let 
X'OX,  the  angle  which  the  cutting  plane  makes  with  the 
plane  of  the  cone's  base,  =  cp. 

Let  P  (x,  y,  z)  be  any  point  on  the  curve  of  intersection 
BPB'.  We  wish  to  find  the  equation  of  this  curve  when 
referred  to  OY,  OX'  as  axes. 

Draw  PD  ||  to  OY ;  PL  and  DK  ||  to  OZ ;  then 

(OK,  KL,  LP)  =  (x,  y,  z)  are  the  space  co-ordinates  of  P, 


252  SOLID  ANALYTIC  GEOMETRY. 

and  (OD,  DP)  =  (x',  y)  are  the  co-ordinates  of  P  when  referred 
to  OX',  OY. 

From  the  figure  KL  =  DP,  PL  =  KD  =  OD  sin  (f,  OK  = 
OD  cos  <jD. 

i.e.,  y  =  y^  z  =  x'  sin  cp,  x  =  x'  cos  (p. 

But  these  values  of  x,  y,  z  must  subsist  together  with  the 
equation  of  the  conic  surface  for  every  point  on  the  curve  of 
intersection;  hence  substituting  in  (2),  Art.  192,  reducing  and 
remembering  tliat  sin^  go  =  cos^  go  tan  ^  (f,  we  have,  dropping 
accents, 

y"'  tan  "  Q  -{-x"^  cos^  (p  (tan  ^  Q  —tan  ^  gp)  +2  ex  sin  go  —  c^  =  0  ...  (1) 

for  the  equation  of  the  intersection. 

By  giving  every  value  to  go  from  0  to  90°  and  to  c  every 
value  from  0  to  oo,  equation  (1)  can  be  made  to  represent  every 
section  cut  from  a  cone  by  a  plane  except  sections  made  by 
planes  that  are  parallel  to  the  co-ordinate  planes. 

Cor.  1.    Comparing  (1)  with  (1),  Art.  138,  we  find 

a  =  tan  ^  6  I 

b=0  l    •  •  •  (2) 

c  =  cos^  go  (tan  ^  0  —  tan  ^  go) 

Hence,  equation  (1)  represents  an  ellipse,  a  parabola,  an 
hyperbola  or  one  of  their  limiting  cases  according  as,  Art. 
146. 

&^  <  4  ac 

P  =  4  ac 

P  >  4:  ac. 

Case  1.  ^  >  go.  We  find  this  supposition  in  (2)  gives  a  >  0 
and  c  >  0 ;  hence,  ^^  <  4  ac,  i.e.,  the  intersection  is  an 
ellipse. 

If  ^  >  gp  and  c  =  0,  the  equation  resulting  from  introducing 
this  supposition  in  (1)  can  only  be  satisfied  by  the  point 
(0,  0)  ;  hence  it  is  the  equation  of  two  imaginary  lines  inter- 
secting at  the  origin. 


THE  STRAIGHT  LINE.  253 

If  (jp  =  0,  equation  (1)  becomes 

y'^  tan  ^  0  -\-  x'  tan  ^  0  =^  c^, 
that  is,  the  intersection  is  a  circle. 

Case  2.    6  =  (p.     This  supposition  in  (2)  gives  a  >  0  and 

G  =  0  .-.  b"  =  4ac.     Hence  the  intersection  is  a  parabola. 
If  ^  =  qo  and  c  =  0.     From  (1),  we  have 
7/'^  tan  '^6  =  0 ;  i.e.,  y  =  0 

which  is  the  equation  of  the  X-axis  —  a  straight  line. 

If  ^  =  go  =  90°  andc  =  gc,  then  the  cone  becomes  a  cylinder, 
and  the  cutting  plane  is  perpendicular  to  its  base.  The  inter- 
section is  therefore  two  parallel  lines. 

Case  3.  6  <C  cp.  This  supposition  makes  a  >  0,  c  <  0 
.*.  5^  >  4  ac.     Hence  the  intersection  is  an  hyperbola. 

It  0  <C  cp  and  c  =  0  then  (1)  becomes 

y^  tan  -  6  =  x^  cos^  cp  (tan  ~  cp  —  tan  ^  6) 

which  is  the  equation  of  two  intersecting  lines. 

Case  4.    Planes  ||  to  the  co-ordinate  planes. 

(a)  Plane  \\  to  XY-jilane.  Let  s  =  m  be  the  equation  of 
such  a  plane.  Combining  it  with  the  equation  of  the  conic 
surface,  we  have 


...   (3) 
tan  2  ^  ^  ^ 

which  is  the  equation  of  a  circle  for  all  values  of  m. 

(b)  Plane  \\  to  YZ-plane.  Let  x  =  n  be  the  equation  of 
such  a  plane.     Combining  with  (2),  Art.  192,  we  have 

(c  _  zf  =  (n'  -4-  7/2)  tan  ^  Q 

or  yH&n  ^  $  -  z^  -]-  2  cz  -i-  nHsin'-  e  -  o''  =  0  .  .  .  (4) 

which,  since  6^  >.  4  ac,  is  the  equation  of  an  hyperbola  for  all 
values  of  n. 

(c)  Plane  \\  to  XZ-plane.  Let  y  =  2^^Q  the  equation  of  such 
a  plane.  Combining  with  (2),  Art.  192,  we  have  after  reduc- 
tion 

xHan'-e  -z^  +  2  cz  +^/  tan  ^  ^  -  c^  =  0  .  .  .  (5) 


254 


SOLID  ANALYTIC  GEOMETRY. 


which,  since  6^  >  4  ac,  is  the  equation  of  an  hyperbola  for  all 
values  of  p. 

Hence,  in  all  possible  positions  of  the  cutting  plane,  the 
intersection  is  an  ellijyse,  a  parabola,  an  hyperbola,  or  one  of 
their  liviiti7ig  cases. 

JSToTE.  —  Equations  (3),  (4),  (5)  of  case  4  are  the  equa- 
tions of  the  projections  of  the  curves  of  intersection  on  the 
planes  to  which  they  are  parallel.  But  the  projection  of  any 
plane  curve  on  a  parallel  plane  is  a  curve  equal  to  the  given 
curve ;  hence  the ,  conclusions  of  case  4  are  true  for  the 
curves  themselves. 

194.  We  have  defined  the  conies,  Art.  191,  as  the  curves 
cut  from  the  surface  of  a  right  circular  cone  by  a  plane,  and 
assuming  this  definition  we  have  found  and  discussed  their 
general  equation.  Art.  193. 

A  conic,  hoivever,  tnay  be  otherwise  defined  as  the  locus 
generated  by  a  point  so  moving  in  a  plane  that  the  ratio  of  its 
distance  from  a  fixed  point  and  a  fixed  line  is  always  constant. 

195-    To  deduce  the  general  equation  of  a  conic. 
Y 


Fig. 


Let  us  assume  the  definition  of  Art.  194  as  the  basis  of  the 
operation.  Let  F  be  the  fixed  point  and  OY  the  fixed  line. 
Let  P  be  the  generating  point  in  any  position  of  its  path. 


THE  STRAIGHT  LINE.  255 

Draw  FO  J_  to  OY,  and  take  OY  and  OX  as  co-ordinate  axes. 
Draw  PL  i|  to  OY,  PD  J_  to  OY,  and  join  P  and  F.     Let  OF 

FP 

By  definition  =  e  =  a  constant. 

From  triangle  FPL,  FP^  =  FL^  +  PL^ ;  .  .  .   (1) 
but  FL2  =  (OL  -  OYf  =  (x  -  j^f,  LP^  =  y'-  ■  and  FP^  = 
e^DP^  =  eV. 

These  values  in  (1)  give 

e-  x^  =  (x  —  iSf  +  xf' 
or,  after  reduction, 

7/2  +  (1  -  (?)  x"  -  Ipx  +^2  ^  0  ...  (2) 
for  the  required  equation. 

CoR.    Comparing  (2)  with  (1),  Art.  138,  we  find 
a  =  1,  5  =  0,  and  c  =  (1  —  e^), 
hence         V  -  ^ac  =  -  ^  (Y  -  (?)  =  ^  {e"  -  V)  .  .  .  (3) 

Case  1.    The  fixed  point  not  on  the  fixed  line ;  i.e.,  p  not 
zero. 

If  e  <  1,  &2  <  4  ac ;  hence  equation  (2)  is  the  equation  of 
an  ellipse. 

If  e  =  1,  6^  =  4  ac ;  hence  equation  (2)  is  the  equation  of  a 
parabola. 

If  e  >  1,  5^  >  4  ac;  hence  equation  (2)  is  the  equation  of 
an  hyperbola. 

Case  2.    The  fixed  point  is  on  the  fixed  line,  i.e.,  ^  =  0. 

In  this  case  (2)  becomes 

y'i  j^  (I  -  e})  x""  =  0  .  .  .  (4) 

If  6  <  1,  equation  (4)  represents  two  imaginary  lines  inter- 
secting at  origin. 

If   6  =  1,  equation    (4)    represents    one    straight  line  (the 
X-axis). 

If  e  >  1,  equation  (4)  represents  two  straight  lines  inter- 
secting at  the  origin. 

Hence,  equation  (2)  represents  the  conies  or  one  of  their  lim- 
iting cases. 


=  2)      X  -2z-\-  3  =  0} 
-If'    2j-z  =  2  \ 


256  SOLID  ANALYTIC   GEOMETRY. 

GENERAL  EXAMPLES. 

1.  Find  the  point  of  intersection  of  the  lines 

x  =  2z  -\-ll      x=z  -{-2 
y  =  3z+2S^     y  =  4.z  +  l 

and  the  cosine  of  the  angle  between  them. 

5 
Ans.     (3,  5,  1)  ;  cos  go  =  — 1| 

2.  Eequired  the  equation  of  the  line  which  passes  through 
(1,  —  2,  3)  and  is  parallel  to 

-  =  2-  +  n.  Ans.    -  =  2.-5    1 

y  =  2-z^  y  =  -«  +  l| 

3.  What  is  the  angle  between  the  lines 

X  -\-z  =  2 

y 

Ans.     (p  =  90°. 

4.  What  is  the  distance  of  the  point  (  —  3,  2,  —  1)   from 

the  line 

ic  +  3«  +  1  ==  0|  ^ 
2/ =  4. +  3         r 

5.  A  line  makes  equal  angles  with  the  co-ordinate  axes ; 
required  the  angles  which  it  makes  with  the  co-ordinate 
planes. 

6.  The  equation  of  a  surface  is  x^  -\-  y"^  -\-  z^  —  2x  —  Ay  — 
6z  =2;  what  does  the  equation  become  when  the  surface  is 
referred  to  a  parallel  system  of  axes,  the  origin  being  at 
(1,  2,  3)  ?  Ans.  x^-\-y^  +  z^  =  16. 

7.  Given  the  line      _  ^  ^"T     [■ ,  required  the  projection  of 

the  line  on  XY  and  the  point  in  which  the  line  pierces  the 
co-ordinate  planes.  Ans.    in  part,  2  3/  +  cc  =  4. 

8.  Eequired  the  distance  cut  off  on  the  Z  and  Y  axes  by 
the  projections  of  the  line  '^  i  o  ^^  o  [-  on  YZ. 

z=  -6 

Ans.  3 

2/  =  o    • 


THE  STRAIGHT  LINE.  257 

9.  How  are  the  following  pair  of  lines  related  ? 

x^2z^2    \      x  =  2z-l    \ 
2/__^_4|      y=-z-\-2\ 

10.  What  are  the  equations  of  the  line  which  passes  through 
the  origin  and  the  point  of  intersection  of  the  lines 

x=2z-^l\      x  =  z  +  2     \^  .  x  =  -Sz\ 

11.  What  is  the  distance  of  the  point  (3,  2,  —  4)  from  the 
origin  ?  What  angle  does  this  line  make  with  its  projection 
on  XY  ? 

12.  A  straight  line  makes  an  angle  of  60°  with  the  X-axis 
and  an  angle  of  45°  with  the  Y-axis ;  what  angle  does  it  make 
with  the  Z-axis  ?  Ans.    60°. 

13.  What  are  the  cosines  of  the  angles  which  the  line 
X  =  3  z  —  1 


,    o  r  makes  with  the  co-ordinate  axes  ? 
y=  —  z-^2 

14.  A  line  passes   through  the   point   (1,  2,  3)  and  makes 

V2     1      1 

angles  with  X,  Y,  Z  whose  cosines  are  — —  >   o '   2  '    respect- 
ively ;  required 

(a)  the  equation  of  the  line, 

(b)  the  equation  of  the  plane  J_  to  the  line  at  the  point, 

(c)  to  show  that  the   projections   of  the  line  are  J_  to  the 
traces  of  the  plane. 

2     12 

15.  The  directional  cosines  of  two  lines  are  - ,  --  ,  ~  and 

o      o      o 

^^,  -,  -.     What  is  the  cosine  of  the  angle    which   they 
2       2     2 

make  with  each  other  ?  _ 

3  -f  2  V2 


Ans.    Cos  <jp  = 


6 


16.  The  projecting  planes  of  a  line  are  x  =  3  z  —  1  and 
X  =  2y  -\-2.  What  is  the  equation  of  the  plane  which  pro- 
jects the  line  on  YZ  ?  A71S.   3  z  —  2  y  =  3. 


258  SOLID  ANALYTIC  GEOMETRY. 

17.  The  projections  of  a  line  on  XZ  and  YZ  each  form  with 
the  Z-axis  an  angle  of  45° ;  required  the  equation  of  the  line 
which  passes  through  (2,  1,  4)  parallel  to  the  line. 

Ans.  H  .  ^ ,  or  „  K 

18.  Pind  the  equation  of  the  line  which  contains  the  point 
(3,  2,  1)  and  meets  the  line      _  ^  _  o     [-at  right  angles. 

19.  Given  the  lines      ~~  ^         -i  r    and       ~  o  ^    ,   ^  [-  :   re- 

y  =  3«  —  1)  ?/  =  3«  +  1  j   ' 

quired 

(ct)  the  value  of  s  in  order  that  the  lines  may  be  parallel ; 

{b)  the  value  of  s  in  order  that  the  lines  may  be  perpen- 
dicular ; 

(c)  the  value  of  s  in  order  that  the  lines  may  intersect. 

2     12 

20.  The  directional  cosines  of  a  line  are  -  ,  -  ,  - ;    required 

o     o     o 

the  sines  of  the  angles  which  the  line  makes  with  the  co-ordi- 
nate planes. 

21.  Find  the  equations  of  the  line  which  passes  through 

the  origin  and  is  perpendicular  to  the  two  lines  ';-     T  o  ^ 

y  —  o  s  -|-  o  ) 

and  ^  =  ^  +  11  ^^^«-   X  =  3z 


y  =  2z       ;•  ^J=-2z 

22.  Find    the    angle    included    between    the   two   planes 
Ax -j- Bij  +  Cz  =  I)  and  A'x  +  B'y  -f  C'z  =  D'. 

AA^  +  BB^  +  CC^ 

V  A2  +  B2  +  C^  VA'2  +  B'^  +  C'-  * 

23.  If  two  planes  are  parallel  show  that  the  coefficients  of 
the  variables  in  their  equations  are  proportional. 

24.  Find   the   condition  for  perpendicularity   of   the  two 
planes  given  in  Example  22. 

A71S.   AA' +  BB' +  CC  =  0. 


SURFACES   OF  THE  SECOND   ORDER.  259 


CHAPTER   IV. 

A  DISCUSSION    OF    THE    SURFACES    OF   THE 
SECOND    ORDER. 

By  A.  L.  Nelson,  M.A.,  Professor  of  Mathematics  in  Washing- 
ton and  Lee  University,  Va. 

Every  equation  involving  three  variables  represents  a  sur- 
face. If  the  equation  be  of  the  first  degree  the  surface  will 
be  a  plane.  If  the  equation  be  of  a  higher  degree  the  surface 
will  be  curved.  It  is  proposed  in  this  chapter  to  determine 
the  nature  of  the  surfaces  represented  by  equations  of  the 
second  degree  involving  three  variables.  The  most  general 
form  of  the  equation  of  the  second  degree  is  Ax^  +  By^  +  Cz^ 
+  Dxy  +  ^xz  +  Et/-  4-  Ga;  +  H^/  +  I«  +  K  =  0  . . .  (1)  where 
the  coefi&cients  A,  B,  C,  etc.,  may  be  of  either  sign  and  of  any 
magnitude.  Let  us  suppose  the  co-ordinate  axes  to  be  rectan- 
gular. The  form  of  equation  (1)  may  be  simplified  by  a 
transformation  of  axes.  Let  us  turn  the  axes  without  chan- 
ging the  origin. 

The  formulae  of  transformation  are  (Art.  190) 

cc  =  cc'  cos  «'  +  y'  cos  a"  -(-  «'  cos  a'" 
y  =  x'  cos  /3'  +  /  cos  ^"  +  .?'  cos  ft'" 
z  =  x'  cos  '('  +  y'  cos  /'  +  z'  cos  Y" 

Substituting  these  values,  equation  (1)  becomes 

A'^'2  ^  ]gy2  ^  c'.^'"  +  V>'x'y'  +  ^x'z'  +  Yy'z'  -f  GV  +  Wy' 
+  IV  +  K  =  0  .  .  .   (2). 


260  SOLID  ANALYTIC  GEOMETRY. 

Since  the  original  axes  were  supposed  rectangular  the  nine 
angles  a',  /3;  y'  etc.,    are  connected  by  the  three  relations 

cos^  «'  +  cos^  /3'  +  cos^  y'  =  1. 
cos2  a"  4-  cos^  ^"  +  cos^  y"  =  1. 

COS^  a'"  +  C0S2  /3'"  +  COS^  f"  =  1. 

If  we  take  the  new  axes  also  rectangular,  which  is  desir- 
able, the  nine  angles  will  be  connected  by  the  three  additional 
relations 

cos  a'  COS  a"'  -{-  cos  ^'  COS  /3"  +  COS  y'  COS  y"  =  0. 

cos  «'  COS  a'"  -|-  COS  (3'  COS  (3'"  +  COS  y'  cos  y"'  =  0. 

COS  «"  cos  «'"  +  cos  (3"  cos  ^'"  +  cos  y"  cos  r'"  =  0. 

This  will  leave  three  of  the  nine  angles  to  be  assumed  arbi- 
trarily.    Let  us  give  to  them  such  values  as  to  render  the  co- 
efficients D',  E',  and  F'  each  equal  to  zero  in  equation  (2). 
The  general  equation  will  thus  be  reduced  to  the  form 
A'x'^  +  By^  +  G'^''  +  G'x'  +  Hy  -f  IV  +  K  =  0, 
or,  omitting  accents, 

Ax^ -\- By'  -\- Gz^  +  Gx -\- mj -\- Iz  +  K  =  0  .  .  .  (3) 
In  order  to  make  a  further  reduction  in  the  form  of  the 
equation  let  us  endeavor  to  move  the  origin  without  changing 
the  direction  of  the  axes.     The  formulae  of  transformation 
will  be  (Art.  189) 

X  =  a  -\-  x',  1/  ^  b  -{-  y',  z  =  c  -\-  z' . 

Equation  (3)  will  become 

A  (a  +  x'Y   +  B   (^<  +  y'Y  +  C  (c  +  z!y  -f  G  (a  +  a?')  + 

H  (Z»  +  y')  +  I  (c  +  ^0  +  K  =  0. 
Developing,  omitting  accents,  and  placing  Ao?  +  BIP-  -\-  Cc^ 
+  Ga  +  HS  +  Ic  +  K  =  L,  the  equation  takes  the  form 
Aa;2  -j-  B^/^  +  Gz"  +  (2  Aa  +  G)  cc  +  (2  Bh  +H)  y  +  (2  Cc  +1  )z 
+  L  =  0. 
In  order  now  to  give  definite  values  to  the  quantities  a,  5, 
and  c,  which  were  entirely  arbitrary,  let  us  assume 

G       (,=  _     H         ,  I     „. 


2A  '  2B    '  20 


SURFACES   OF  THE  SECOND   ORDER.  261 

2  Aa  +  G  =  0,  2  B6  +  H  =  0,  2  Cc  +  I  =  0  .  .  .  (4) 
If  these  values  of  a,  b,  and  c  be  finite,  the  general  equation 
reduces  to  the  form 

A^2  +  By^  +  C^^  +  L  =  0  .  .  .  [A], 

a  form  which  will  be  set  aside  for  further  examination. 

It  may  be  remarked  that  equations  (4)  are  of  the  first 
degree,  and  will  give  only  one  value  to  each  of  the  quantities 
a,  b,  and  c,  and  there  is  therefore  only  one  position  for  the 
new  origin. 

If,  however,  either  A,  B,  or  C  be  zero,  then  a,  b,  or  c  will 
become  infinite,  and  the  origin  will  be  removed  to  an  infinite 
distance.     This  must  be  avoided. 

Let  us  suppose  A  =  0,  while  B  and  C  are  finite.  We  may 
then  assume  2  B6  +  H  =  0,  and  2  Cc  +  I  =  0,  but  we  cannot 
assume  2  A.a  -|-  G  =  0. 

Having  assumed  the  values  of  b  and  c  as  indicated,  let  us 

assume  the  entire  constant  term  equal  to  zero.     This  will  give 

B^»2  4_  Cc2  +  Ga  +  H6  +  Ic  +  K  =  0, 

B&2  +  Cc2  +  H5  +  Ic  +  K 

or  a  = ! ! ! 1 

G 

and  the  general  equation  will  be  reduced  to  the  form 
By'' +  Cz^  +  Gx  =  0  .  .  .  (B), 

a  second  form  set  aside  for  examination. 

We  must  observe  that  this  last  proposed  transformation 
will  also  fail  when  G  =  0,  that  is,  when  the  first  power  of  x, 
as  well  as  the  second  power  of  x,  is  wanting  in  the  general 
equation. 

And  without  making  the  second  transformation  we  have  a 
third  form  for  examination,  viz. : 

By2  +  c^^  +  Hy  +  I^  +  K  =  0  .  .   .   (C) 

Lastly,  two  of  the  terms  involving  the  second  powers  of 
the  variables  may  be  wanting,  and  the  equation  (1)  then 
becomes 

C^2  +  Gx  +  Hy  +  I;s  +  K  =  0  .  .  .  (D) 


262  SOLID  ANALYTIC   GEOMETRY. 

It  is  apparent,  therefore,  that  every  equation  of  the  second 
degree  involving  three  variables  can  be  reduced  to  one  or 
another  of  the  four  forms 

Kx"  +  B?/2  +  C«-  +  L  =  0  .  .  .  (A) 
B?/2  +  C,^^  +  G^  =  0  .  .  .  (B) 
B2/2  +  C«2  +  Hy  +  I«  +  K  =  0  .  .  .  (C) 
C«2  +  Gx  +  Hy  +  I^  +  K  =  0  .  .  .   (D) 

We  will  examine  each  of  these  forms  in  order,  beginning 
with  the  first  form  : 

Aa;2  +  B7/2  +  C^2  _^  L  =  0  .  .  .  (A) 

This  equation  admits  of  several  varieties  of  form  according 
to  the  signs  of  the  coefficients. 

1.  A,  B,  and  C  positive,  and  L  negative  in  the  first  member. 

2.  A,  B,  C,  and  L  positive. 

3.  Two  of  the  coefficients  as  A  and  B  positive,  C  and  L 
negative. 

4.  Two  of  the  coefficients  as  A  and  B  positive,  C  negative, 
and  L  positive. 

No  other  cases  will  occur. 

Case  1.  Kx"  +  By'-  +  Cs^  =  L, 

in  which  form  all  of  the  coefficients  are  positive. 

In  order  to  determine  the  nature  of  the  surface  represented 
by  this  equation,  let  it  be  intersected  by  systems  of  planes 
parallel  respectively  to  the  co-ordinate  planes.  The  equations 
of  these  intersecting  planes  will  he  x  =  a,  y  ^=  b,  z  =  c.  Com- 
bining the  equations  of  these  planes  with  that  of  the  surface, 
we  find  the  equations  of  the  projections  on  the  co-ordinate 
planes  of  the  curves  of  intersection. 


When  X  =  a, 

■By^  +  C«2  =  L  -  Aa^ 

an  ellipse. 

"    y  =  h, 

Ax2  +  C«2  =  L  -  BP 

an  ellipse. 

"     z  =  c, 

Ax^  +  By'  =L-  Cc2 

an  ellipse. 

Thus  we  see  that  the  sections  parallel  to  each  of  the  co- 
ordinate planes  are  ellipses. 


SURFACES  OF  THE  SECOND   ORDER. 


263 


The    section    made    by   the    plane    x  =  a    is    real    when 
L  —  A(x-  >  0  or  a  <  -j-  y— r-j  and  imaginary  in  the  contrary 

case. 

The    section    made    by   the    plane    y  =  h    is    real    when 

ft  <  J^  t/ — ,  and  imaginary  when  ^  >  J^  V/^~  • 

The    section    made    by    the    plane    2:  =  c    is    real   when 

c  <  -t  y  -j;- ,  and  imaginary  when  c  >  i  W  — -  . 

Thus  we  see  that  the  surface  is  enclosed  within  a  rectangu- 
lar parallelepiped  whose  dimensions  are 


When        a 


B 


and 


ovb  =  ^ 


\    —  or  c  = 
V  B 


± 


the 


A''       ^VB  -^VC 

sections  become  points. 

When  a  =  0,h  =  0,  and  c  =  0,  we  find  the  sections  made, 
by  the  co-ordinate  planes  to  be 

B^/'  +  Cz^  =  L. 

Kx^  +  C«2  =  L. 

Xx-  +  B?/-  =  L. 

E 


264  SOLID  ANALYTIC  GEOMETRY. 

These  are  called  the  principal  sections  of  the  surface.  The 
principal  sections  are  larger  ellipses  than  the  sections  parallel 
to  them,  as  is  indicated  by  the  magnitude  of  the  absolute 
term. 

The  surface  is  called  the  Ellipsoid. 

It  may  be  generated  by  the  motion  of  an  ellipse  of  variable 
dimensions  whose  centre  remains  on  a  fixed  line,  and  whose 
plane  remains  always  perpendicular  to  that  line,  and  whose 
semi-axes  are  the  ordinates  of  two  ellipses  Avhich  have  the 
same  transverse  axis,  but  unequal  conjugate  axes  placed  at 
right  angles  to  each  other.  The  axes  of  the  principal  sections 
are  called  the  axes  of  the  ellipsoid. 

If  we  represent  the  semi-axes  of  the  ellipsoid  by  a,  J,  and 
c,  we  shall  have 

and  the  equation  of  the  surface 

kx'  +  B?/-  +  Cs-  =  L  becomes 

r^  7/2  z^ 

-^  +  i^  +  -^  =  1,  or 

a2    ^    ^2    ^    ^2 

Tr(?x^  -f-  o^c^'if'  -\-  a-lrz'  =  a}lrc^. 

These  are  the  forms  in  which  the  equation  of  the  ellipsoid 
is  usually  given. 

If  we  suppose  B  =  A,  then  b  =  a,  and  the  equation  becomes 

^•'  +  2/"    I    «'  _  1 
cr  c- 

and  the  surface  is  the  ElUi^soid  of  Bevohdion  about  the  axis 

of  Z. 

If  A  =  B  =  C,  then  a  =  h  =  c,  and  the  equation  becomes 

^2  _|_  yi  j^  ^ji  ^  (.^2^  and  the  surface  is  a  sphere. 


If  L  =  0,  the  axes  2  sj^,  2  y/-^,  2  sj ^ 
reduce  to  zero,  and  the  ellipsoid  becomes  a  point. 


SURFACES  OF  THE  SECOND   ORDER.  265 

Case  2.  If  L  be  negative  in  the  second  member,  the  equa- 
tion A.x^  +  B?/^  +  Cs;^  =  —  L  will  represent  an  imaginary 
surface,  and  there  will  be  no  geometrical  locus. 

Hence  the  varieties  of  the  elUjosoid  are 

(1)  The  ellipsoid  proper  with  three  unequal  axes. 

(2)  The  ellipsoid  of  revolution  with  two  equal  axes. 

(3)  The  sphere. 

(4)  The  point. 

(5)  The  imaginary  surface. 

Case  3.    In  this  case  the  equation  takes  the  form 
Ax^  +  B2/2  _  C^2  ^  L, 

in  which  A,  B,  C,  and  L  are  essentially  positive. 

Cutting  the  surface  by  planes  as  before,  the  sections  will  be, 
when  cc  =  a,    By^  —  Cz^  =  L  —  Ao-^,  a  hyperbola,  having  its 

transverse  axis  parallel  to  the  Y-axis  when  «,<_[_  t  / —  ,  but 
parallel  to  the  Z-axis  when  a  >  -[-  i  / And  when  a  =  J- 

— - ,  the  intersection  becomes  two  straight  lines  whose  pro- 

jections  on  the  plane  of  YZ  pass  through  the  origin. 

When  y  =  b,  Kx"^  —  Gz^  =  L  —  B5^,  a  hyperbola,  with  simi- 
lar conditions  as  above. 

When  z  =  c,  Kx"^  +  Bi/  =  L  +  Cc^,  an  ellipse  real  for  all 
values  of  c. 

Since  the  elliptical  sections  are  all  real,  the  surface  is  con- 
tinuous, or  it  consists  of  a  single  sheet. 

The  principal  sections  are  found  by  making  successively 
a  =  0,  which  gives  B?/^  —  Cz^  =  L,  a  hyperbola. 
5  =  0,       "  "     Ax''  -  Cz^  =  L,  « 

c  =  0,       "  "      Aa;2  -f  By^  =  L,  an  ellipse. 

The  surface  is  called  the  elliptical  hyperholoid  of  one  sheet. 
The  equation  may  be  reduced  to  the  form 
x^    .    y'^         z'^   _  -, 
a?-         y         c^ 


/ 


266 


SOLID  ANALYTIC  GEOMETRY. 


This  surface  may  be  generated  by  an  ellipse  of  variable 
dimensions  whose  centre  remains  constantly  on  the  Z-axis, 
and  whose  plane  is  perpendicular  to  that  axis,  and  whose 
semi-axes  are  the  ordinates  of  two  hyperbolas  having  the 
same  conjugate  axis  coinciding  with  the  Z-axis,  but  having 
different  transverse  axes  placed  at  right  angles  to  each  other. 


Fig.  B. 


If  we  suppose  A  =  B,  then  will  a  =  h,  and  the  equation  of 
the  surface  becomes 


9       1  ^ 


-  — =  1, 


the  hyperboloid  of  revolution  of  one  sheet. 

If  A  =  B  =  C,  we  have  x""  -^  y"^  —  z"  =  a^,  the  equilateral 
hyperboloid  of  revolution  of  one  sheet. 

If  L  =  0,  the  equation  represents  a  right  cone  having  an 
elliptical  base ;  and  if  A  =  B  this  base  becomes  a  circle. 


SURFACES  OF  THE  SECOND   ORDER.  267 

Hence  we  have  the  following  varieties  of  the  hyxjerholoid 
of  one  sheet. 

1.  The  hyperboloid  proper,  with  three  unequal  axes. 

2.  The  hyperboloid  of  revolution. 

3.  The  equilateral  hyperboloid  of  revolution. 

4.  The  cone. 

Case  4.  A.x''  +  Bif  -  Gz'' =  -  L,  where  A,  B,  C,  and  L 
are  essentially  positive. 

Intersecting  the  surface  as  before  we  have,  when  x  =  a, 
B,r/2  _  Cs;2  =^  —  L  —  Ka^,  a  hyperbola  having  its  transverse 
axis  parallel  to  the  axis  of  Z. 

When  y  =  h,  A-x"  —  C«'  =  -  L  -B^^^  A  hyperbola  hav- 
ing its  transverse  axis  parallel  to  the  axis  of  Z. 

When  ^_=  c,  Ax^  +  By^  =  —  L  +  Cc^,  an  ellipse  real  when 

c  >  -i;  i/ — .,  and  imaginary  when  c  <  -^  y  ~7r'      Since  the 

sections  between  the  limits  «  =  J-  y —  are  imaginary,  but 

real  beyond  those  limits,  it  follows  that  there  are  two  distinct 
sheets  entirely  separated  from  each  other. 

The  surface  is  called  the  hyperboloid  of  two  sheets. 

The  principal  sections  are  found  by  making  successively 

a  =  0,  which  gives  By^  —  Cs;^  =  —  L,  a  hyperbola  with  its 
transverse  axis  coinciding  with  the  Z-axis. 

b  =0,  which  gives  Ax^  —  Cz^  =  —Jj,  a,  hyperbola  with  its 
transverse  axis  coinciding  with  the  Z-axis. 

c  =  0,  which  gives  Ax^  -f-  By^  =  —  L,  an  imaginary  ellipse. 

The   semi-axes  of   the   first   section   are   i/jz_andt/_. 

V  B  V  C 

Those  of  the  second  section  are  1/ and  i/— .     And  those 


A  V    C 


of   the   imaginary   section    are   y — .  (— 1)  and  i  / (— !)• 

The  distances  2  i/       ,  2  v/-:^ ,  and  2  1/-^-  are  called  the  axes 


268 


SOLID  ANALYTIC   GEOMETRY. 


Fig.  C. 


of  the  surface.     Eepresenting  the  semi-axes  by  a,  b,  and  c, 
the  equation  of  the  surface  may  be  reduced  to  the  form 

„9      "1"       70  o 


If   we   suppose  A  =  B,  then  a  =  b,  and  the  equation  re- 
duces to 

x^  +  f  _  ^'  _       1 
the  hyperboloid  of  revolution  of  two  sheets. 


SURFACES  OF  THE  SECOND  ORDER.  269 

If  A  =  B  =  C,  the  equation  becomes 
x^  -\-  if  —  z^  =  —  a-, 
which  represents  tlie  surface  generated  by  the  revokition  of 
an  equilateral  hyperbola  about  its  transverse  axis. 

Finally,  if  L  =  0,  the  surface  becomes  a  cone  having  an 
elliptical  base,  and  the  base  becomes  a  circle  when  A  =  B. 

We  have,  therefore,  the  following  varieties  of  the  liyper- 
boloid  of  two  sheets  : 

1.  The  hyperboloid  proper  having  three  unequal  axes. 

2.  The  hyperboloid  of  revolution. 

3.  The  equilateral  hyperboloid  of  revolution. 

4.  The  cone. 

We  will  now  examine  the  second  form, 

B2/2  +  C.~2  +  G  a;  =  0  .  .  .  (B) 

Three   cases  apparently   different  present   themselves   for 
examination. 

(1).    B  and  C  positive  and  G  negative  in   the  first  mem- 
ber. 

(2).    B,  C,  and  G  positive. 

(3).    B  positive  and  C  and  G  negative. 

Case  1.    The  equation  may  be  written 
B7/2  +  C«2  =  Ga; 
in  which  B,  C,  and  G  are  essentially  positive. 

Let  the  surface  be  intersected  as  usual  by  planes  parallel 
respectively  to  the  co-ordinate  planes. 

When  X  =  a,  B?/^  +  Cs^  =  Ga,  an  ellipse  real  when  a  >  0, 
and  imaginary  when  a  <  0. 

When   y  =  b,  Cz'^  =  Gx  —  B6^,    a   parabola   with   its   axis 
parallel  to  the  axis  of  X. 

When   z  =  c,  B?/^  =  Gx  —  Cc^    a   parabola    with   its    axis 
parallel  to  the  axis  of  X. 

The  principal  sections  are  found  by  making  a  =  0,b  =  0, 
and  c  =  0. 

When  a  =0,  B?/^  +  Gz^  =  0,  a  point,  the  origin. 

When  b  =  0,  Gz'^  =  Ga:;,  a  parabola  with   its  vertex  at  the 
origin. 


270 


SOLID  ANALYTIC  GEOMETRY. 


When  c,  =  0,  Bt/^  =  Gx,  a  parabola  with  vertex  at  the  origin. 

Since  every  positive  value  of  x  gives  a  real  section,  and 
every  negative  value  of  x  an  imaginary  section,  the  surface 
consists  of  a  single  sheet  extending  indefinitely  and  contin- 
uously in  the  direction  of  positive  abscissas,  but  having  no 
points  in  the  opposite  direction  from  the  origin. 

The  surface  is  called  the  elliptical  paraboloid.  It  may  be 
generated  by  the  motion  of  an  ellipse  of  variable  dimensions 
whose  centre  remains  constantly  on  the  same  straight  line, 
and  whose  plane  continues  perpendicular  to  that  line,  and 
whose  semi-axes  are  the  ordinates  of  two  parabolas  having  a 
common  axis  and  the  same  vertex,  but  different  parameters 
placed  with  their  planes  perpendicular  to  each  other. 


Fig.  D. 


Case  2.    If  we  suppose  G  to  be  positive  in  the  first  member 
so  that  the  equation  will  take  the  form 

B7/2  4-  C^-  =  -  Gx, 


SURFACES  OF  THE  SECOND   ORDER.  271 

the  sections  perpendicular  to  the  X-axis  will  become  imaginary 
when  X  >  0,  and  real  when  a;  <  0. 

In  other  respects  the  results  are  similar  to  those  deduced  in 
case  1. 

Thus  the  equation  will  represent  a  surface  of  the  same  form 
as  in  case  1,  but  turned  in  the  opposite  direction  from  the 
co-ordinate  plane  of  YZ. 

If  B  =  C,  the  surface  becomes  the  paraboloid  of  revolution. 

Case  3.  By^  _  C^^  =  Go?. 

Intersect  the  surface  by  planes  as  before. 

When  X  ^=  a,  B?/^  —  Gz^  =  Ga,  a  hyperbola  with  transverse 
axis  in  the  direction  of  the  Y-axis  when  a  >  0,  and  in  the 
direction  of  the  Z-axis  when  a.  <  0. 

When  y  =  b,  Gz^  =  —  Qx  -\-  B6^,  a  parabola  having  its  axis 
in  the  direction  of  the  X-axis  and  extending  to  the  left. 

When  z  =  c,  By^  =  Gx  +  Cc^,  a  parabola  having  its  axis  in 
the  direction  of  the  X-axis  and  extending  to  the  right. 

Since  every  value  of  x,  either  positive  or  negative,  gives  a 
real  section,  the  surface  consists  of  a  single  sheet  extending 
indefinitely  to  the  right  and  left  of  the  plane  of  YZ.  This 
surface  is  called  the  Hyperbolic  Paraboloid.  To  find  its  princi- 
pal sections  make  x,  y,  and  z  alternately  equal  to  zero. 

When  X  =  0,  By'^  =  Cz^,  two  straight  lines. 

When  y  =  0,  Cz~  =  —  Gx,  a  parabola  with  axis  to  the  left. 

When  z  =  0,  By^  =  Gx,  a  parabola  with  axis  to  the  right. 

The  hyperbolic  paraboloid  admits  of  no  variety. 

Now  taking  up  form  (C),  By""  +  Cs;^  +  H?/  +  I«  +  K  ^  0, 
we  see  that  it  is  the  equation  of  a  cylinder  whose  elements 
are  perpendicular  to  the  plane  of  YZ,  and  whose  base  in  the 
plane  of  YZ  will  be  an  ellipse  or  hyperbola  according  to 
the  signs  of  B  and  C. 

The  fourth  form  (D),  Cz"  ^  Gx -{- B.y  +  Iz -\- K  =  0  repre- 
sents a  cylinder  having  its  bases  in  the  planes  XZ  and  YZ 


272 


SOLID  ANALYTIC  GEOMETRY, 


Fig.  E. 


parabolas,  and  having  its  right-lined  elements  parallel  to  the 
plane  XY  and  to  each  other,  but  oblique  to  the  axes  of  X  and  Y. 

The  preceding  discussion  shows  that  every  equation  of  the 
second  degree  between  three  variables  represents  one  or  an- 
other of  the  following  surfaces  : 

1.  The  ellipsoid  with  its  varieties,  viz. ;  the  ellipsoid 
proper,  the  ellipsoid  of  revolution,  the  sphere,  the  point, 
and  the  imaginary  surface. 


SURFACES   OF  THE  SECOND   ORDER. 


273 


2.  The  hyperboloid  of  one  or  two  sheets,  witli  their 
varieties,  viz. :  the  hyperboloid  proper  of  one  or  two  sheets, 
the  hyperboloid  of  revolution  of  one  or  two  sheets,  the 
equilateral  hyperboloid  of  revolution  of  one  or  two  sheets, 
the  cone  with  an  elliptical  or  circular  base. 

3.  The  paraboloid,  either  elliptical  or  hyperbolic,  with  the 
variety,  the  paraboloid  of  revolution. 

4.  The  cylinder,  having  its  base  either  an  ellipse,  hyper- 
bola, or  parabola. 

Surfaces  of  Revolution.  —  The  general  equatioii  of  surfaces 
of  revolution  may  be  deduced  by  a  direct  method,  as  follows  : 


Fig.  F. 


Let  the  Z-axis  be  the  axis  of  revolution,  and  let  the  equa- 
tion of  AB,  the  generating  curve  in  the  plane  of  XZ,  be 
x^  =  fz. 

Let  P  be  the  point  in  this  curve  which  generates  the  circle 


274  SOLID  ANALYTIC   GEOMETRY. 

PQR,  and  let  r  be  the  radius  of  the  circle.     We  will  have 
r^  z=  x^  -\-  y^. 

The  value  of  r  may  also  be  expressed  in  terms  of  z  from 
the  equation  of  the  generatrix  in  the  plane  of  XZ  as  follows : 

r2^CF^=  Od'  =fz. 
Equating  these  two  values  of  r  we  have 

x^  +  y""  =fa 

as  the  general  equation  of  surfaces  of  revolution. 

It  will  be  observed  that  the  second  value  of  r'^  is  the  value 
of  x^  in  the  equation  of  the  generatrix.  Hence,  to  find  the 
equation  of  the  surface  of  revolution  we  have  only  to  substi- 
tute x^  +  'if  of  the  surface  for  x^  in  the  generatrix. 

Surface  of  a  Sjihere.  —  Equation  of  generatrix  x^  -\-  z"^  =  W: 
Hence  the  equation  of  the  surface  of  the  sphere  is 

x^J^ifJ^z''  =  ^K 

Ellipsoid  of  Revolution.  — 

Generatrix      —  +  —  =  1. 
a"         c^ 

Surface  ^!J:J^  +  ^  =  i. 

a^  & 

Similarly,  the  equation  of  the  hyperboloid  of  revolution  is 

x^  -\-  y-  _   z'   _  ^ 

a^  c- 

Paraholoid  of  Revolution.  — 

x"^  =  'ipz,  the  generatrix. 

x^  -{-  y-  =  4,2jz,  the  surface  of  revolution. 
Cone  of  revolution,     z  =  mx  -\-  /3  the  generatrix, 


7Ji      '  m? 


Hence  x"^  +  y- 


_(z-Jl 


or  m^  (x^  +  y-)  =  (z  —  0)- 


SURFACES  OF  THE   SECOND   ORDER.  275 

EXAMPLES. 

1.  What  is  the  locus  in  space  of  4  a;^  +  9  y^  =  35  ?  Qf 
9  ^2  —  16  ?/-  =  144  ?  Of  a;'  +  2/'  =  r^  ?  Of  y'  +  -'  =  r^  ?  Of 
y2  +  8  a;  =  0  ? 

2.  Determine  the  nature  of  the  surfaces  a;^  +  2/^  +  4  «^  =  25, 

3.  Find  the  equation  of  the  surface  of  revolution  about  the 
axis  of  Z  whose  generatrix  is  s  =  3  a;  +  5. 

4.  Find  the  equation  of  the  cone  of  revolution  Avhose  inter- 
section with  the  plane  of  XY  is  a;^  +  y^  =  9,  and  whose  vertex 
is  (0,  0,  5.) 

5.  Determine  the  surfaces  represented  by 

ic2  ^  4  2/'  +  9  ^2  ^  36. 
a;2  _|.  4  ^2  _  9  ^2  ^  3e_ 

a;2  +  4  ?/'  =  9  «2  _  36. 
4  3/2  _|_  9  ^2  _  36  ^_ 

4  ?/2  -  9  «2  =  36  a;. 


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